I like the double arrow space, as a classical example:
Let $X = [0,1] \times \{0, 1\}$, where $X$ has the lexicographical ordering $(x,i) < (y,j)$ iff $x < y$ or ( $x = y$ and $i=0, j=1$). Then $X$ in the order topology is separable ($\mathbb{Q} \times \{0, 1\}$ is countable and dense), compact, hereditarily normal and perfectly normal and first countable, but its square is not hereditarily normal (it contains the square of the Sorgenfrey line, which can be seen as the subspace $(0,1) \times \{1\}$ ).
So even very nice compact spaces need not be metrizable.
Proofs can be found here, e.g.
The lexicographically ordered unit square, also discussed in the previous link, is another example, which is less nice (not separable), but for which it is easier to disprove metrizability, as it's compact and not separable.
Another classical example from the same Aleksandrov paper IIRC, is the double of $[0,1]$, which is also $X = [0,1] \times \{0,1\}$, and where a basic neighbourhood of $(x,1)$ is just $\{(x,1)\}$ (these are isolated points), but a basic neighbourhood of $(x,0)$ is of the form $O = (I \times \{0,1\}) \setminus \{(x,1)\}$
where $I$ is any standard open set of $[0,1]$ containing $x$. This is compact as $[0,1]$ is, but has an uncountable discrete open subspace $[0,1] \times \{1\}$, making it not separable and not second countable, so not metrisable.
Since $X$ is second countable, all you need for $X^*$ to be second countable is a countable neighborhood base for the point at infinity, right? Since $X$ is locally compact and second countable, you can cover $X$ with countably many open sets whose closures are compact. Then every compact subset of $X$ is contained in a finite union of those open sets, and the complements of the closures of those unions will provide a countable neighborhood base for $\infty$.
Best Answer
Let $X$ be the real line with the topology in which the usual open sets are open, and in addition $U\setminus A$ is open for any $U$ that is open in the usual topology, where $A=\{\frac1n : n=1,2,...\}$. In other words, every point except the origin has its usual neighborhoods, and the basic neighborhoods of the origin are of the form $(-\varepsilon,\varepsilon)\setminus A$. Then $A$ is a closed subspace that cannot be separated by disjoint neighborhoods from the origin, so $X$ is not regular, and not metrizable. $X$ is Hausdorff since its topology is stronger than the usual topology which is Hausdorff. If we take a countable basis for the usual topology together with a countable local basis at the origin for the new topology, then these two together form a countable basis for the new topology. (This is a standard example, e.g. Ex.1.5.6 in General Topology by R. Engelking.)