Extension of continuous map in topological space In the book  Simmons, George F., Introduction to topology and modern analysis,  page no- 98, question no- 2, the problem is : Let $X$ be a topological space and a $Y$ be metric space and $f:A\subset X\rightarrow Y$ be a continuous map. Then $f$ can be extended in at most one way to a continuous mapping of $\bar{A}$ into $Y$.
I am trying to prove this way. Let $x_0\in \bar{A}-A$ and suppose that there is two extension $f$ and $g$ such that $f(x)=g(x)$ for $x\in A$. Now $f(x_0)\in \overline{f(A)}$ and $g(x_0)\in\overline {g(A)}$. So there exists a sequence $\{f(x_n)\}$ and $\{g(y_n)\}$ that converge to $f(x_0)$ and $g(x_0)$ respectively, where $x_n$ and $y_n$ belong to $A$ for all $n$. Then I am stuck!! Please help to complete the proof.

Some caveats: If A is also a metric space, then f must be uniformly continuous, as this preserves Cauchy sequences , while standard continuity does not.
An example, for $A =\mathbb Q \subset \mathbb R$ , choose $f(x):= \frac {1}{ x-\sqrt 3} $. This is continuous in $\mathbb Q$ but will not extend continuously into $\mathbb R$, as $f$ will blow up near $\sqrt 3$. So you can't extend , since $f$ is not uniformly continuous and will not preserve Cauchy, i.e., while the sequences $\{x_n\}$ with $x_n \rightarrow x$ are Cauchy, the "Pushforward " sequences {$f(x_n) \}$ with $x_n \rightarrow x$ are not, and will therefore not necessarily converge.
