Set of Hölder continuous functions is compact

Question

Let $$H=\{x\in C[0,1]: \forall s,t\in [0,1]: |x(s)-x(t)|\leq b|s-t|^\alpha\}$$ where $C[0,1]$ are the continuous functions on $[0,1]$ with $x(0)=0$ and $\alpha\in (0,1/2)$. Take a sequence $(x_n)\in H$. Choosing $\delta<(\frac{\varepsilon}{b})^{1/\alpha}$, shows $\{x_n\}$ shows the family is equicontinuous. Furthermore $|x(t)|\leq bt^\alpha$ so the family is bounded and by Arzela Ascoli its precompact. So there exists a subsequence, also denoted $x_n$ converging to some $x_0$ uniformly. For all $s,t\in [0,1]$ we have \begin{align*} |x_0(s)-x_0(t)|&\leq |x_0(s)-x_n(s)|+|x_0(t)-x_n(t)|+|x_n(s)-x_n(t)| \\ &\leq |x_0(s)-x_n(s)|+|x_0(t)-x_n(t)|+b|s-t|^\alpha \\ &\to b|s-t|^\alpha. \end{align*} so $x\in H$. So $H$ is compact subset of $C[0,1]$. This has to be wrong right?