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.

To follow your idea: Suppose we have $f_1$ and $f_2$ that are both continuous extensions of $f: A \to Y$ to $\overline{A}$. Let $p \in \overline{A}$ and so we have a net $a_i, i \in (I,\le)$ from $A$ such that $a_i \to p$.
The continuity of $f_1$ implies that $f_1(a_i) \to f_1(p)$.
The continuity of $f_2$ implies that $f_2(a_i) \to f_2(p)$ .
But the nets (as $f_1(a_i) = f_2(a_i) = f(a_i)$) are identical so must converge to the same point in the metric space $Y$ (limits of sequences are unique).
We conclude that $f_1(p) = f_2(p)$
As this holds for all $p \in \overline{A}$, $f_1 = f_2$ on $\overline{A}$.
