We will utilize a proof by contradiction, where we assume that there are two functions $h$ and $g$ such that there exists some $x \ \in \ \bar{A}$ such that $g(x) \ \neq \ h(x)$, and derive a contradiction, showing that $g \ = \ h$, and the extension of $f$ is unique.

We map $x$ under $g$ and $h$ to get $g(x), \ h(x) \ \in \ Y$. Since $g(x) \ \neq \ h(x)$, and $Y$ is Hausdorff, there exists open set $U_g$ and $U_h$ around $g(x)$ and $h(x)$ repsectively such that: $$U_g \ \cap \ U_h \ = \ \emptyset$$ Since $g$ and $h$ are both continuous, it follows that $h^{-1}(U_h)$ and $g^{-1}(U_g)$ are open in $\bar{A}$ and both contain $x$. It follows that their intersection is also an open set, $U$, containing $x$.

Now, we know that $x \ \in \ \bar{A}$, so it follows that every neighbourhood of $x$ intersects $A$. Thus, it follows that $U \ \cap \ A \ \neq \ \emptyset$. In the case that $x \ \in \ A$, by definition of an extension of a function, $g(x) \ = \ h(x) \ = \ f(x)$, which is a contradiction to the initial assumption.

In the case that $x \ \notin \ A$, to follows that $U$ must contain some other point $y \ \in \ A$. Again, by definition of an extension of a function, $g(y) \ = \ h(y) \ = \ f(y)$. Since $y \ \in \ h^{-1}(U_h)$ and $y \ \in \ g^{-1}(U_g)$, it follows that $h(y) \ = \ f(y) \ \in \ U_h$ and $g(y) \ = \ f(y) \ \in \ U_g$. This implies that: $$f(y) \ \in \ U_g \ \cap \ U_h$$ which contradicts the fact that $U_g$ and $U_h$ are disjoint. Thus, it follows that that for all $x \ \in \ \bar{A}$, we have $g(x) \ = \ h(x)$, and the two functions are equal, meaning that $g : \bar{A} \ \rightarrow \ Y$ is uniquely determined by $f$.