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, 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.

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, 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. A neighbourhood base for points $(x,y)$ with $y > 0$ is consists of open discs of radius $< y$. A neighbourhood base for points $(x,0)$ consists of sets of the form $\{ (x,0) \} \cup B_r(x,r)$ for $0 < r < \infty$, where $B_r(x,r)$ is the open disc of radius $r$ with centre $(x,r)$.

The rational points contained in the open upper half-plane form a countable dense set, so this is a separable space. The subspace topology on the horizontal axis is discrete. 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.