Corollary

If $f$ is a finite continuous real-valued function defined on $A$, then there exists a finite continuous real-valued function $g$ defined on $X$, which extends $f$.

First consider the case in which $f(x) \geq 0$ for all $x \in A$; then there is a continuous extension $g_1$ of $f$ to $X$, taking its values in $[0, +\infty]$. If we put $B = \overline{g_1}(+\infty)$, then $B$ is closed and by hypothesis does not meet $A$; the function $h$ which is equal to $f$ on $A$ and to 0 on $B$ is therefore a continuous function on the closed set $A \cup B$. Let $g_2$ be a continuous extension of $h$ to $X$, again taking its values in $[0, +\infty]$; the function $g = \inf(g_1, g_2)$ is then a continuous extension of $f$ to $X$, whose values are $\geq 0$ and finite at every point of $X$.

To pass to the general case, it is enough to remark that, if $f$ is finite and continuous on $A$, then so are $f^+$ and $f^-$; extending $f^+$ and $f^-$ to $X$ by finite continuous functions $g_1$ and $g_2$ respectively, we see that the function $g_1 - g_2$ is finite and continuous on $X$ and extends $f$.