Working on the following problem from Munkres:
Let $(X, d_{X})$ and $(Y, d_Y)$ be metric spaces; let $Y$ be complete. Let $A \subset X$. Show that if $f:A \to Y$ is uniformly continuous, then $f$ can be uniquely extended t a continuous function $g: \overline{A} \to Y$, and $g$ is uniformly continuous.
If there exists such an extension, it must be unique, and $f(x) = \lim_{n \to \infty}f(x_n)$ for $x_n$ a sequence in $A$ converging to $x \in \overline{A}$. To show this limit exists, let $x_n \to x$, then $f(x_n)$ is Cauchy, since uniformly continuous functions take Cauchy sequences to Cauchy sequences. Then $f(x_n)$ converges by completeness, and I claim it converges to $f(x)$.
Now, to show $g$ is uniformly continuous, I kind of brute-forced it, and I was wondering if anyone had a more elegant method. (I consider the proof up to now pretty elegant.) What I did was: Suppose for contradiction that $\exists \epsilon >0$ s.t. $\forall \delta >0\,\, \exists x_\delta, y_\delta \in \overline{A}$ s.t. $d(x_\delta, y_\delta)< \delta$ and $d(g(x_\delta), g(y_\delta))> \epsilon$. Let $\delta := \delta_{\epsilon}/2$, where $\delta_\epsilon$ is taken from the uniform continuity of $f$. Now $d(g(\cdot), g(\cdot))$ is continuous...in particular, it is continuous in each variable separately. Therefore we can vary $x_{\delta}$ and $y_{\delta}$ a little bit, to get $x^*,\, y^* \in A$ such that $d(x^*, y^*)<\delta_{\epsilon}$ and $d(f(x^*), f(y^*))>\epsilon$, a contradiction.
Does anyone know a more elegant method to show the uniform continuity?