We have that if $X$ is a metric space, $Y$ is a complete metric space, $A\subseteq X$ is dense in $X$ and $f:A \to Y$ is uniformly continuous then there exists a (unique) uniformly continuous extension of $f$ to $X$. This has been asked here many times, my question regards a variation of this result.
Does this still hold if $A$ is not necessarily dense in $X$? That is, for every $A\subseteq X$ and every uniformly continuous $f:A\to Y $ there exists at least one uniformly continuous extension to $X$.
I believe this is not true, and I think I've been able to find a counterexample, but it's much more complicated than what would seem necessary given the question, and I'd like to know if there's some obvious counterexample that I may have missed.
Sure, consider $X$ to be any connected space (e.g. $\mathbb{R}$), pick two distinct points $x,y\in X$ and let $A=Y=\{x,y\}$. Of course a finite metric space is trivially complete.
With that we have a very simple counterexample. Let $f:A\to Y$ be $f(x)=x$. Clearly it cannot be extended to full $X$, even in continuous way, because $X$ is connected while $Y$ is not.
Note that if $Y=\mathbb{R}^n$ (and even more generally I think it holds for any ANR), and $f:A\to Y$ is uniformly continuous and bounded, then $f$ can be extended to uniformly continuous and still bounded $\overline{f}:\overline{A}\to Y$, and then it can be extended to full $X$ (in uniformly continuous way), as a variant of Tietze extension theorem.
I don't know whether "bounded" condition can be dropped. Obviously $Y$ cannot be replaced with arbitrary complete space.