Consider the $B$-space $X = C([0, 1])$ equipped with the norm $\Vert · \Vert_{\infty}$. And let $F$ be a closed subspace of $X$ such that every function $u\in F$ is Holder continuous, namely, for each $u \in F$ there exist $\alpha \in (0, 1]$ and $L\gt0$ such that $ |u(t) − u(s)| \le L|t − s|^{\alpha} \ \ \forall t,s \in [0,1] $.
$(i)$ Setting, for every $n\in \mathbb{N}_{0}$ \begin{align} F_{n} = \{ u \in F : |u(t) − u(s)| \le n|t − s|^{1/n}, \ \ \forall t, s \in [0, 1] \} \end{align} and using the Baire theorem prove that there exist $ \beta \in (0, 1]$ and $M\gt0 \ $ (both independent of $u$) such that, for all $ u\in F$ \begin{align} |u(t) − u(s)| \le M \Vert u \Vert_{\infty}|t − s|^{\beta}, \ \ \forall t, s \in [0, 1] \end{align}
$(ii)$ Prove that the closed unit ball $B_{F} \subset F$ centered at $0$ is compact and deduce that F is finite-dimensional.
If I have a compact ball in $F$, doesn't it contradict the fact that in a infinite-dim space ($X$ in this case) balls are never compact?
No. $\alpha$-Hölder is not preserved by $\lVert-\rVert_\infty$, so $F$ is not a closed subspace of $(C([0,1]), \lVert-\rVert_\infty)$ (e.g., $F$ contains all polynomials, so its closure in sup norm is $C([0,1])$ by Stone-Weierstrass). See also this question from a few hours ago.