Let $F\subseteq C^1([0,1],\mathbb{R})$ be a closed subspace of $C([0,1],\mathbb{R})$. Show that $F$ is of finite dimension.
So, I considered the norm $\Vert f\Vert_1=\Vert f\Vert_\infty+\Vert f'\Vert_\infty$ and showed that $(F,\Vert\cdot\Vert_1)$ is a Banach space. Next, using Arzela-Ascoli, I showed that $\overline{\mathbb{B}}_{(F,\Vert\cdot\Vert_1)}(0,1)$ is compact in $(F,\Vert\cdot\Vert_\infty)$. At this point, I got help and was told to show that $\Vert\cdot\Vert_\infty$ and $\Vert\cdot\Vert_1$ are equivalent norms on $F$ and that I could use the open mapping theorem to do so... How?
The next step would be to use the Riesz compactness theorem to conclude that $F$ is, in fact, of finite dimension. But, in which step, exactly, did we need the closedness of $F$? I have the feeling of not having talked about it explicitly.
You have $\|x\|_{\infty}\leq \|x\|_1$, this implies that $Id:(C^1([0,1]),,\|\|_1)\rightarrow (C^1([0,1]),\|\|_{\infty}$ is continuous (a sequence which converges for $\|\|_1$ converges for $\|\|_{\infty}$), since it is surjective, it is an homemorphism (open mapping), and $Id^{-1}(B_{\|\|_{\infty}}(0,1))=B_{\|\|_{\infty}}(0,1)$ is open thus there exists $c>0$ such that $B_{\|\|_{1}}(0,c)\subset B_{\|\|_{\infty})}(0,1)$ and the norms are equivalent.
Hint: since $\|\|_{\infty}$ is equivalent to $\|\|_{1}$, $Id:C^1([0,1])\rightarrow C^1([0,1])$ is an homeomorphism, you deduce that $B_{F,\|\|_1}(0,1)$ is compact and $F$ is finite dimensional. You have to review the proof that $(F,\|\|_1)$ is Banach to see if you used the fact that $F$ is closed.