Is the extension of a continuous function on a subspace to the closure of its domain unique? The following problem is from Munkres's Topology (exercise 13 in section 18 "Continuous Functions"; page 112, 2nd edition).

Let $A \subset X$; let $f:A \to Y$ be continuous; let $Y$ be a Hausdorff. Show that if $f$ may be extended to a continuous function $g:cl(A)\to  Y$, then $g$ is uniquely determined by $f$.

 A: HINT: Suppose that $g_0:\operatorname{cl}A\to Y$ and $g_1:\operatorname{cl}A\to Y$ both extend $f$. We know that $g_0(x)=g_1(x)$ for each $x\in A$, so suppose that $x\in(\operatorname{cl}A)\setminus A$. For convenience let $y_0=g_0(x)$ and $y_1=g_1(x)$. Suppose that $y_0\ne y_1$; $Y$ is Hausdorff, so there are open sets $U_0$ and $U_1$ in $Y$ such that $y_0\in U_0$, $y_1\in U_1$, and $U_0\cap U_1=\varnothing$. 

Note: Everything that I’ve done up to here is just ‘following my nose’, i.e., doing the most natural, straightforward thing that might work; it’s the sort of preliminary reasoning that should quickly become nearly automatic. For instance, if you want to prove that something is unique, one obvious approach is to assume that it isn’t unique and try to get a contradiction; that’s what I’m doing here. That gives me the distinct points $y_0$ and $y_1$; now what is there in the statement of the problem that naturally goes with having a pair of distinct points in $Y$? Here that’s easy. The hypothesis that $Y$ is Hausdorff immediately tells me something about those two points: they can be separated by disjoint open nbhds, and I might as well do it and see whether it’s useful.

And here’s the rest of the hint: let $V_0=g_0^{-1}[U_0]$ and $V_1=g_1^{-1}[U_1]$, and consider the sets $V_0\cap A$ and $V_1\cap A$ to get a contradiction.
