It is well known, that every subspace of separable metric space is separable. It is also known, this statement not to be true, if space is topological and not necessary metric. But I cannot find an example of topological uncountable and non-metrizable space and topology $\tau$ is infinite, such that every subspace is still separable.
Example of “almost metric” topological space
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A nice example is the Sorgenfrey line, which is hereditarily separable, uncountable, Hausdorff and hereditarily normal, and first countable, so it has much in common with metrizable spaces, but since it is separable and not second countable, it is not metrizable.
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Every second countable space is hereditarily separable, since second countability is hereditary and implies separability.
Here is a list of uncountable second-countable, nonmetrisable spaces: link. None of them are $T_3$, however, since a second-countable $T_3$ space is metrisable by the Urysohn metrisation theorem, as pointed out by @HelloDarkness in the comments.
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If a space is separable, then it remains separable under any coarser topology (any dense subset remains dense if you make the topology coarser). So, you could take your favorite uncountable separable metric space (say, $\mathbb{R}$) and take a coarser topology that is not metrizable. Of course the easiest way to do this is to take the indiscrete topology, but you ask for the topology to be infinite to avoid such trivialities presumably. For a less trivial example, you could take the cofinite topology. Or, for instance, you could take the topology on $\mathbb{R}$ consisting of just intervals of the form $(-x,x)$ (and the empty set). There are many other similar examples you can come up with in this way.
Take the line with two origins. It is second countable so every subspace is second countable. Second countable implies separable. But it is not Hausdorff, so is not metrizable.