Every closed subspace of ${\scr C}^0[a,b]$ of continuously differentiable functions must have finite dimension. If $F \subset {\scr C}^1[a,b] \subset {\scr C}^0[a,b]$, then $\dim F < +\infty,$ where $F$ is a closed subspace (in $ {\scr C}^0[a,b]$). I found this answer, which is very good and solves the problem, but how can we prove the assertion without using Ascoli-Arzelà in the end? We didn't cover equicontinuity, etc. 
Is there a more elementary proof of the result?
 A: The inclusion $\mathscr{C}^1[a,b] \hookrightarrow \mathscr{C}^0[a,b]$ is continuous, so $F$ is also closed in $\mathscr{C}^1[a,b]$. Thus $F$ is a Banach space with respect to the $\mathscr{C}^0$-norm and with respect to the $\mathscr{C}^1$-norm. Hence these two norms are equivalent on $F$ (since they are comparable). Say we have
$$\lVert f\rVert_{\mathscr{C}^1} \leqslant K\cdot \lVert f\rVert_{\mathscr{C}^0}\tag{1}$$
for all $f\in F$.
Let $n > K\cdot (b-a)$. Consider the subspace
$$F_n = \Bigl\{ f \in F : f\bigl(a + k\tfrac{b-a}{n}\bigr) = 0,\text{ for } 0 \leqslant k \leqslant n\Bigr\}.$$
$F_n$ is the null space of a (continuous) linear map $F \to\mathbb{C}^{n+1}$, hence $F_n$ has finite codimension ($\leqslant n+1$) in $F$.
Suppose $F_n \neq \{0\}$, i.e. there were an $f \in F_n$ with $\lVert f\rVert_{\mathscr{C}^0} = 1$. Say the maximal modulus $1$ is attained at $t\in [a,b]$. The distance from $t$ to the next point $a + k\frac{b-a}{n}$ is at most $\frac{b-a}{2n}$. By the mean value theorem, we therefore have
$$\lvert f'(x)\rvert \geqslant \frac{2n}{b-a} > 2K$$
for some $x\in [a,b]$, but that means
$$\lVert f\rVert_{\mathscr{C}^1} > 2K \lVert f\rVert_{\mathscr{C}^0}.$$
This contradicts $(1)$, hence there is no $f \in F_n$ with $\lVert f\rVert_{\mathscr{C}^0} = 1$, i.e. $F_n = \{0\}$, and consequently $\dim F \leqslant n+1$.
