A Hausdorff space ${\displaystyle X\,}$ is paracompact if and only if it every open cover admits a subordinate partition of unity. The if direction is straightforward. Now for the only if direction, we do this in a few stages.

Lemma 1: If ${\displaystyle {\mathcal {O}}\,}$ is a locally finite open cover, then there exists open sets ${\displaystyle W_{U}\,}$ for each ${\displaystyle U\in {\mathcal {O}}\,}$, such that each ${\displaystyle {\bar {W_{U}}}\subseteq U\,}$ and ${\displaystyle \{W_{U}:U\in {\mathcal {O}}\}\,}$ is a locally finite refinement.

Lemma 2: If ${\displaystyle {\mathcal {O}}\,}$ is a locally finite open cover, then there are continuous functions ${\displaystyle f_{U}:X\to [0,1]\,}$ such that ${\displaystyle \operatorname {supp} ~f_{U}\subseteq U\,}$ and such that ${\displaystyle f:=\sum _{U\in {\mathcal {O}}}f_{U}\,}$ is a continuous function which is always non-zero and finite.

Theorem: In a paracompact Hausdorff space ${\displaystyle X\,}$, if ${\displaystyle {\mathcal {O}}\,}$ is an open cover, then there exists a partition of unity subordinate to it.

Proof (Lemma 1):

Let ${\displaystyle {\mathcal {V}}\,}$ be the collection of open sets meeting only finitely many sets in ${\displaystyle {\mathcal {O}}\,}$, and whose closure is contained in a set in ${\displaystyle {\mathcal {O}}}$. One can check as an exercise that this provides an open refinement, since paracompact Hausdorff spaces are regular, and since ${\displaystyle {\mathcal {O}}\,}$ is locally finite. Now replace ${\displaystyle {\mathcal {V}}\,}$ by a locally finite open refinement. One can easily check that each set in this refinement has the same property as that which characterised the original cover.

Now we define ${\displaystyle W_{U}=\bigcup \{A\in {\mathcal {V}}:{\bar {A}}\subseteq U\}\,}$. The property of ${\displaystyle {\mathcal {V}}\,}$ guarantees that every ${\displaystyle A\in {\mathcal {V}}}$ is contained in some ${\displaystyle W_{U}}$. Therefore ${\displaystyle \{W_{U}:U\in {\mathcal {O}}\}\,}$ is an open refinement of ${\displaystyle {\mathcal {O}}\,}$. Since we have ${\displaystyle W_{U}\subseteq U}$, this cover is immediately locally finite.

Now we want to show that each ${\displaystyle {\bar {W_{U}}}\subseteq U\,}$. For every ${\displaystyle x\notin U}$, we will prove that ${\displaystyle x\notin {\bar {W_{U}}}}$. Since we chose ${\displaystyle {\mathcal {V}}}$ to be locally finite, there is a neighbourhood ${\displaystyle V[x]}$ of ${\displaystyle x}$ such that only finitely many sets in ${\displaystyle {\mathcal {V}}}$ have non-empty intersection with ${\displaystyle V[x]}$, and we note ${\displaystyle A_{1},...,A_{n},...\in {\mathcal {V}}}$ those in the definition of ${\displaystyle W_{U}}$. Therefore we can decompose ${\displaystyle W_{U}}$ in two parts: ${\displaystyle A_{1},...,A_{n}\in {\mathcal {V}}}$ who intersect ${\displaystyle V[x]}$, and the rest ${\displaystyle A\in {\mathcal {V}}}$ who don't, which means that they are contained in the closed set ${\displaystyle C:=X\setminus V[x]}$. We now have ${\displaystyle {\bar {W_{U}}}\subseteq {\bar {A_{1}}}\cup ...\cup {\bar {A_{n}}}\cup C}$. Since ${\displaystyle {\bar {A_{i}}}\subseteq U}$ and ${\displaystyle x\notin U}$, we have ${\displaystyle x\notin {\bar {A_{i}}}}$ for every ${\displaystyle i}$. And since ${\displaystyle C}$ is the complement of a neighbourhood of ${\displaystyle x}$, ${\displaystyle x}$ is also not in ${\displaystyle C}$. Therefore we have ${\displaystyle x\notin {\bar {W_{U}}}}$.

${\displaystyle \blacksquare \,}$ (Lem 1)

Proof (Lemma 2):

Applying Lemma 1, let ${\displaystyle f_{U}:X\to [0,1]\,}$ be continuous maps with ${\displaystyle f_{U}\upharpoonright {\bar {W}}_{U}=1\,}$ and ${\displaystyle \operatorname {supp} ~f_{U}\subseteq U\,}$ (by Urysohn's lemma for disjoint closed sets in normal spaces, which a paracompact Hausdorff space is). Note by the support of a function, we here mean the points not mapping to zero (and not the closure of this set). To show that ${\displaystyle f=\sum _{U\in {\mathcal {O}}}f_{U}\,}$ is always finite and non-zero, take ${\displaystyle x\in X\,}$, and let ${\displaystyle N\,}$ a neighbourhood of ${\displaystyle x\,}$ meeting only finitely many sets in ${\displaystyle {\mathcal {O}}\,}$; thus ${\displaystyle x\,}$ belongs to only finitely many sets in ${\displaystyle {\mathcal {O}}\,}$; thus ${\displaystyle f_{U}(x)=0\,}$ for all but finitely many ${\displaystyle U\,}$; moreover ${\displaystyle x\in W_{U}\,}$ for some ${\displaystyle U\,}$, thus ${\displaystyle f_{U}(x)=1\,}$; so ${\displaystyle f(x)\,}$ is finite and ${\displaystyle \geq 1\,}$. To establish continuity, take ${\displaystyle x,N\,}$ as before, and let ${\displaystyle S=\{U\in {\mathcal {O}}:N{\text{ meets }}U\}\,}$, which is finite; then ${\displaystyle f\upharpoonright N=\sum _{U\in S}f_{U}\upharpoonright N\,}$, which is a continuous function; hence the preimage under ${\displaystyle f\,}$ of a neighbourhood of ${\displaystyle f(x)\,}$ will be a neighbourhood of ${\displaystyle x\,}$.

${\displaystyle \blacksquare \,}$ (Lem 2)

Proof (Theorem):

Take ${\displaystyle {\mathcal {O}}^{*}\,}$ a locally finite subcover of the refinement cover: ${\displaystyle \{V{\text{ open }}:(\exists {U\in {\mathcal {O}}}){\bar {V}}\subseteq U\}\,}$. Applying Lemma 2, we obtain continuous functions ${\displaystyle f_{W}:X\to [0,1]\,}$ with ${\displaystyle \operatorname {supp} ~f_{W}\subseteq W\,}$ (thus the usual closed version of the support is contained in some ${\displaystyle U\in {\mathcal {O}}\,}$, for each ${\displaystyle W\in {\mathcal {O}}^{*}\,}$; for which their sum constitutes a continuous function which is always finite non-zero (hence ${\displaystyle 1/f\,}$ is continuous positive, finite-valued). So replacing each ${\displaystyle f_{W}\,}$ by ${\displaystyle f_{W}/f\,}$, we have now — all things remaining the same — that their sum is everywhere ${\displaystyle 1\,}$. Finally for ${\displaystyle x\in X\,}$, letting ${\displaystyle N\,}$ be a neighbourhood of ${\displaystyle x\,}$ meeting only finitely many sets in ${\displaystyle {\mathcal {O}}^{*}\,}$, we have ${\displaystyle f_{W}\upharpoonright N=0\,}$ for all but finitely many ${\displaystyle W\in {\mathcal {O}}^{*}\,}$ since each ${\displaystyle \operatorname {supp} ~f_{W}\subseteq W\,}$. Thus we have a partition of unity subordinate to the original open cover.

${\displaystyle \blacksquare \,}$ (Thm)