If $X \times Y$ is Separable, are $X, Y$ Separable?

If $X = \emptyset$ and $Y$ is not separable, then $X \times Y = \emptyset$ is separable. However, this counterexample along with the symmetric counterexample where $Y = \emptyset$ are the only possible counterexamples.

So, suppose $X$ and $Y$ are both nonempty, and $D = \{ (x_n, y_n) \mid n \in \mathbb{N} \}$ is dense in $X \times Y$. We then claim that $\pi_1(D) = \{ x_n \mid n \in \mathbb{N} \}$ is dense in $X$. In fact, suppose we have any nonempty open subset $U \subseteq X$. Then since $Y$ is nonempty, $\pi_1^{-1}(U)$ is a nonempty open subset of $X \times Y$. Since $D$ is dense, this implies that $(x_n, y_n) \in \pi_1^{-1}(U)$ for some $n$, which means that $x_n \in U$. The proof that $\pi_2(D) = \{ y_n \mid n \in \mathbb{N} \}$ is dense in $Y$ is similar.

Yes, as $X = \pi_1[X \times Y]$ and the continuous image of a separable space is separable (if $f: X \to Y$ is continuous and surjective and $D$ is dense in $X$: $$Y = f[X] = f[\overline{D}] \subseteq \overline{f[D]}$$ so that $f[D]$ is dense in $Y$ ; the inclusion follows from the continuity.)

So $X \times Y$ separable implies $X$ separable, and using the other projection, $Y$ is separable too. If both are separable with countable dense subsets $D_X$ resp. $D_Y$, then $D_X \times D_Y$ is countable and dense in $X \times Y$, and so $X \times Y$ is separable. So we have equivalence.

For second countability we have a similar proof except that for $X \times Y$ second countable implies $X$ is, we use that $X$ embeds as a subspace in the product, and so does $Y$.