I am currently self-studying Munkres Topology, and I wanted to check the correctness of my proof for the following problem. There are definitely more concise proofs online (like this one), but I want to make sure my logic is sound, especially in the first three sentences of the second paragraph. Any critiques are welcome!
The Problem:
Munkres 18.13: Let $A\subset X$; let $f:A\rightarrow Y$ be continuous; let $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g:\bar A\rightarrow Y$, then $g$ is uniquely determined by $f$.
My Attempt: Let $h:\bar A\rightarrow Y$ be another continuous extension of $f$, and assume to the contrary that $h(x)\ne g(x)$ for some $x\in \bar A$. Let $U,V$ be two disjoint neighborhoods of $g(x),h(x)$ respectively, which exist by $Y$ being Hausdorff. Then $g^{-1}(U)\cap h^{-1}(V)\subset \bar A$ is open in $\bar A$ by the continuity of $g$ and $h$, implying this set is given by $\bar A \cap O$ for some $O$ open in $X$.
We see $g^{-1}(U)\cap h^{-1}(V)$ contains $x$. Then $O$ is a neighborhood of $x$ in $X$, meaning $O$ must intersect $A$ as $x$ is in the closure of $A$. This implies $g^{-1}(U) \cap h^{-1}(V)$ contains an element of $A$, since $g^{-1}(U)\cap h^{-1}(V) = \bar A \cap O$. Denoting such an element by $a$, we then have $f(a) = g(a)\in U$ and $f(a)=h(a)\in V$. This shows $U$ and $V$ are not disjoint, a contradiction. Therefore $h = g$. $\diamond$
