Where is the use of continuity in this Munkres Topology question? 

  
*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 $f$.
  

I believe I've solved this, here's an outline:
I assume that for $x \in \overline{A} - A$ we have $g(x) \neq h(x)$ for $g,h$ extensions of $f$ as assumed.
Using the Hausdorff assumption I take two disjoint environments $V_g, V_h \subset Y$.
We have that $x \in g^{-1}(V_g) \cap h^{-1}(V_h) := U$ so $U$ is an environment of $x$ in $X$.
By assumption $\exists z \in A \cap U$ which implies that $f(z) = h(z) = g(z)$ in contradiction to $V_g, V_h$ being disjoint.
Where did I use the continuity of $f$, or is this proof wrong?

You're taking in $U$ the intersection of two inverse images of open sets under $g$ and $h$. The latter are open by continuity of $g$ resp. $h$ (the intersection is open by the topology axioms). The openness is needed for it to intersect $A$.
The continuity of $f$ is needed because if $f$ has a continuous extension $g$ to $\overline{A}$,  then $g|A$ is also continuous by a standard result in subspaces, and this is by definition $f$ again.
We don't say that $f$ has a continuous extension (it need not have: extend $f(x) = \frac{1}{x}$ to all of the reals..), but if there is one, there can be only one and not more.
