Let $X$ be a metrizable subset of $[0,1]^I$ for some uncountable set $I$. Suppose that $f\colon X\to [0,1]$ is a uniformly continuous function with respect to some metric that metrizes $X$. Can we extend $f$ to a continuous function on the compact set $\overline{X}$?
The problem is that $\overline{X}$ need not be metrizable anymore, so the standard proof does not work.
Not necessarily. Indeed, first of all we remark that every continuous function $f$ from $X$ to $[0,1]$ is uniformly continuous with respect to some metric $d$ that metrizes $X$. For instance, pick any metric $d’$ that metrizes $X$ and put $d(x,y)=d’(x,y)+|f(x)-f(y)|$ for each $x,y\in X$. Now we recall Corollary 3.6.3 from Engelking's “General Topology” (2nd ed., Heldermann, Berlin, 1989). Every continuous function from a Tychonoff space $X$ to the closed interval $I$ is extendable to a continuous function from the Čech-Stone compactification of $X$. If every continuous function from a Tychonoff space $X$ to the closed interval $I$ is continuously extendable over a compactification $\alpha X$ of $X$, then $\alpha X$ is equivalent to the Čech-Stone compactification of $X$. As a concrete counterexample we may pick as $X$ any convergent sequence in $[0,1]^I$ (with distinct members but without its limit).