The answer is no, if we interpret separable as "has a countable dense subset" (Fedor Petrov's answer appears to have interpreted it as possessing a countable base). Consider the [Moore plane][1], or Niemytzki tangent disc topology, as it is called on page 100 of Steen & Seebach's *Counterexamples in Topology*. This is a topology on the closed upper half-plane.

The rational points in the open upper half-plane form a countable dense set, so this is a separable space. The horizontal axis has the discrete topoology as a subspace. Now, the discrete topology is always metrizable, and as the real axis has continuum cardinality and every set is equal to its closure, it is *not* separable. So we have a non-separable metrizable subspace of a separable space.


  [1]: https://en.wikipedia.org/wiki/Moore_plane