Problem with extension of a continous function 
Let $X$ a first-countable Topological Space, let $Y$ an Hausdorff Topological Space, let $A\subset X$ a subset ot $X$ and let $f:A\rightarrow Y$ a continous function.
Prove that, if there is an extension
$$\overline{f} :\overline{A}\rightarrow Y$$
$\overline{f}$ is solely determined from $f$.

I thought that:
If there is $g$ that is another extension of $f$, and I call $Z=\lbrace x\in \overline{A}\mid \overline{f} (x)=g(x)\rbrace$. Then, $Y$ is an Hausdorff Space so, $Z$ is closed in $X$ and,  $A$ is dense in $\overline{A}$ so $A\subseteq Z\Rightarrow \overline{A}\subseteq Z\Rightarrow \overline{A} = Z$
But I don't think it's right because I didn't use the fact that $X$ is first-countable. I have to use also the fact that $f$ is continous so if  $x_{n}\rightarrow x\Rightarrow f(x_{n})\rightarrow f(x)$.
Can someone help me?

Your argument works. You may need to elaborate a bit on

Then, $Y$ is a Hausdorff space so, $Z$ is closed in $X$

and it may be preferable to write $Z$ is closed in $\overline{A}$ there. Whether you need to elaborate or not depends on what properties of Hausdorff spaces and continuous maps can be assumed as generally known.
First countability of $X$ plays no role in it, it was probably assumed to enable arguments with sequences for those who aren't yet used to topological arguments using open and closed sets, neighbourhoods, preimages etc.
I find your argument much preferable to a sequence (or net/filter for the version not assuming first countability) argument.
