Brezis says that for a bounded open interval $I$, the injection $W^{1,p}(I) \subset L^\infty(I)$ is continuous for all $1 \leq p \leq \infty$, the injection $W^{1,p}(I) \subset C(\bar{I})$ is compact for $1 < p \leq \infty$.
Does this imply the injection $W^{1,p}(I) \subset L^\infty(I)$ is also compact? The two issues are that the first compact injection is into $C(\bar{I})$, not $C(I)$, and that the injection is not into $L^\infty(I)$.
Since the values of a continuous function on an open set determine the values on the closure, restriction to $I$ provides a natural injection $C(\bar I) \hookrightarrow L^\infty (I)$. Since we use the sup norm for both spaces, this map is continuous. Composing it with the compact embedding $W^{1,p} \hookrightarrow C(\bar I)$ gives a compact embedding $W^{1,p} \hookrightarrow L^\infty (I)$.