Let $$K = \{f \in L^2 ([0, 1]) \,\, | \,\, \|f\|_{L^2} \leq 1 \} $$$$ L = \{f:[0, 1] \to \mathbb C \,\, | \,\, |f(x) - f(y)| \leq |x - y| \,\, \forall x, y \in [0, 1]\} $$ The question is to show $ K \cap L $ is compact in $ L^2 ([0, 1]) $.
I have shown that both $ K $ and $ L $ are closed in $ L^2 ([0, 1]) $, and that $ K $ is not compact in $ L^2 ([0, 1]) $. As $ L^2 ([0, 1]) $ is complete in its norm, it would suffice to show that $ K \cap L $ is totally bounded, but I'm not sure how to begin proving this.
The set $K \cap L$ is uniformly bounded by $2$:
Indeed, assume that $f \in K \cap L$ is such that $\left|f(t_0)\right| > 2$ for some $t \in [0,1]$. Then for any $t \in [0,1]$ we have $$2 - \left|f(t)\right|< \left|f(t_0)\right| - \left|f(t)\right|\le \left|f(t_0) - f(t)\right| \le |t_0 - t|$$
Hence $\left|f(t)\right| > 2 - |t_0 - t| \ge 1$. Now we would have $\int_0^1|f(t)|^2\,dt > 1$ which is a contradiction with $f \in K$.
The set $K \cap L$ is equicontinuous. This is clear because all functions are $1$-Lipschitz.
Arzelà–Ascoli implies that every sequence in $K \cap L$ has a uniformly convergent subsequence, so it also converges in $L^2$.
We conclude that $K \cap L$ is compact.