# Show that there exists at most one extension of $f$ whose co-domain is a Hausdorff space [duplicate]

I want to show the following

Suppose $A \subset X, f: A \to Y$ is continuous, $Y$ is Hausdorff. Show that there is at most one continuous extension $g: \overline A \to Y$

I feel like I am very close but I cannot finish.

By way of contradiction, suppose that there exists another extension $h: \overline A \to Y$, take $x \in \overline A \backslash A$, we claim that $h(x) \neq g(x)$

Since $Y$ is Hausdorff, there exists open sets $U, V, U \cap V = \varnothing$ such that $g(x) \in U$, and $h(x) \in V$

Then $x \in g^{-1}(U) \cap h^{-1}(v) \subset \overline A$

How do I proceed from here?

Since $g$ and $h$ are continuous, $g^{-1}(U)\cap h^{-1}(V)$ is an open set containing $x$ and hence it contains an element $a$ of $A$. But $a\in g^{-1}(U)\cap h^{-1}(V)$ implies that $f(a)=g(a)\in U$ and $f(a)=h(a)\in V$ which contradicts the assumption that $U,V$ are disjoint