[proofplan] A [Banach space](/page/Banach%20Space) carries a canonical metric induced by its norm. Since every separable [metrizable space](/page/Metrizable%20Space) has the property that all of its subspaces are separable ([Subspaces of Separable Metrizable Spaces](/theorems/942)), the closed subspace $Y$ — being in particular a subspace of the separable metrizable space $X$ — is separable. [/proofplan] [step:Observe that the norm on $X$ induces a metric, making $X$ a metrizable topological space] Define \begin{align*} d: X \times X &\to [0, \infty) \\ (x, y) &\mapsto \|x - y\|_X. \end{align*} This is a metric on $X$: it satisfies $d(x,y) = 0$ if and only if $x = y$ (definiteness of the norm), $d(x,y) = d(y,x)$ (since $\|x - y\|_X = \|y - x\|_X$), and the triangle inequality $d(x,z) \le d(x,y) + d(y,z)$ (from $\|x - z\|_X = \|(x - y) + (y - z)\|_X \le \|x - y\|_X + \|y - z\|_X$). The [topology](/page/Topology) induced by $d$ is the norm topology on $X$, so $X$ is a [metrizable](/page/Metrizable%20Space) topological space. [/step] [step:Apply the subspace separability theorem for metrizable spaces to conclude $Y$ is separable] The space $X$ is separable by hypothesis and [metrizable](/page/Metrizable%20Space) by the preceding step. The closed subspace $Y \subset X$ is, in particular, a subspace of $X$ (the closedness hypothesis is stronger than what is needed for this step). By [Subspaces of Separable Metrizable Spaces](/theorems/942), every subspace of a separable metrizable space is separable. Therefore $Y$ is separable. [/step]