Suppose that $X$ is a compact connected metric space, $A$ is a dense subset of $X$, and $K$ is a compact subset of $X$. $A$ does not have to be dense in $K$ (for example take $K$ to be the rationals in $[0,1]$, $A$ to be the irrationals in $[0,1]$ and $X$ to be $[0,1]$ itself).
What conditions guarantee that $A$ is dense in $K$? Is it enough to require that $K$ be connected also?
This is true when $K$ is a regular closed set, so $K = \overline{\mathrm{int}(K)}$. It is standard that $A$ is dense in $\mathrm{int}(K)$ and so $A$ is also dense in its closure $K$.
For non regular closed $K$ things can easily fail: $X = [0,4]$, $A$ the rationals in $[0,4]$, $K = [0,\frac12] \cup \{\pi\}$, then $\pi$ is not in the closure of $A \cap K$.