Since $cX$ is compact Hausdorff, it is normal, so two closed sets $A,B \subset cX$ are disjoint iff there is a continuous $f : cX \to [0,1]$ which separates them (i.e. $f=0$ on $A$ and $f=1$ on $B$). Such $f$ is the extension of a bounded uniformly continuous $g : X \to [0,1]$.
Can you find two disjoint closed subsets of $X$ which cannot be separated by a bounded uniformly continuous function?
Their closures in $cX$ will be as desired.
For the second part, following my comment, you could find a sequence in $cX$ that has no convergent subsequence, showing that $cX$ is not sequentially compact. $cX$ is compact by construction, and by the Bolzano-Weierstrass theorem, every compact metrizable space is sequentially compact, so this would rule out metrizability.
Maybe we can even choose this sequence to lie in $X$.
Suppose we have a sequence $\{y_n\} \subset X$ that converges in $cX$. In particular, for any continuous $f : cX \to [0,1]$, $\lim f(y_n)$ exists. As before, any bounded uniformly continuous $g : X \to [0,1]$ extends to a continuous function on $cX$, so $\lim g(y_n)$ must exist as well.
So I propose the following:
Find a sequence $\{x_n\} \subset X$ such that, for every subsequence $\{x_{n_k}\}$, there is a bounded uniformly continuous $g: X \to [0,1]$ such that $\lim g(x_{n_k})$ does not exist.
This sequence will have no convergent subsequence in $cX$.