$X^*$ is separable then $X$ is separable [Proof explanation]

Take a bounded linear functional $f \in Y^{\ast}$ s.t. $f\vert_Y = 0$. Then choose some $u \in X \setminus Y$ with $\Vert u \Vert = 1$ (which exists since $Y$ is proper), define $g \in (Y \oplus \mathbb{C}u)^{\ast}$ via $g(u) = 1$ and $g\vert_Y = f$ and extend linearly.

Then we have $\Vert g \Vert = 1$, since for every $x \in X$ with $\Vert x \Vert = 1$ we either have $g(x) = 1$ or $g(x) = 0$. Hence Hahn-Banach gives an extension $G \in X^{\ast}$ s.t. $\Vert G \Vert = \Vert g \Vert = 1$ and $G \vert_Y = g \vert_Y = f = 0$.

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The above uses the norm preserving version of Hahn-Banach. But really, you only need any extension of the above defined $g$. Then you can you may rescale, and since $g\vert_Y = 0$, the rescaled $g$ is also $0$ on $Y$. The sublinear functional used for this would be $x \mapsto \Vert g \Vert \cdot \Vert x \Vert$.