Prove or disprove: $(Lip((a,b)),\lVert\cdot \rVert_{\infty})$ is separable.
My attempt: Since in a metric space any subspace of a separabale space is separable, and $(\mathcal{C}^0([a,b]),\lVert\cdot \rVert_{\infty})$ is separable (and it is obviously a metric space), we conclude $(Lip((a,b),\lVert\cdot \rVert_{\infty})$ is separable.
I had some doubts about this proof becuase I've read that, in general, only open subspaces of a separable space are separable: but since in this contest we're dealing with metric spaces, which have more structure, any subspace of a separable space is separable.
Thus I'd like to understand if my proof works fine and what I've written in the second paragraph is true. Thanks in advance.
The space is separable. Take a specific dense set of $C^{0}([a,b])$, polynomials, and simply point out that these live in Lip([a,b]) because they are $C^{1}([a,b])$ and so have bounded first derivatives (on $[a,b]$). The proof (if it exists) would have to be different if you allowed infinite intervals.