$L^1$ is complete as a normed space, while $L^2$ is complete as an inner-product space.Suggest a slogan
Let \((X, \mcm, \mu)\) be a measure space.
\(L^1(X,\mu)\), with the norm \(\norm{f}_1 \coloneqq \int_X \abs{f}\dmu\), is a Banach space.
\(L^2(X,\mu)\), with the inner product \(\inner{f}{g} \coloneqq \int_X f\,\overline{g}\dmu\), is a Hilbert space.
\(L^2(X,\mu)\) need not be separable. Let \(X\) be an uncountable set, \(\mcm\) its power set, and \(\mu\) counting measure. The indicator functions \(\chi_{\theset{x}}\) for \(x\in X\) lie in \(L^2(X,\mu)\) and satisfy \(\norm{\chi_{\theset{x}}-\chi_{\theset{y}}}_2=\sqrt2\) for \(x\neq y\), so the open balls of radius \(\sqrt2/2\) about them are pairwise disjoint and uncountably many; a countable dense subset would meet each of them.