Finite dimensional subspace of $C([0,1])$ 
Let linear $S$ be a subspace of $C([0,1])$, i.e., the continuous real-valued
  functions on $[0,1]$. Assume that there exists $c>0$, such that
  $\|\,f\|_\infty\leq c \|\,f\|_2$, for all $f\in S$. Then show that $S$ is finite-dimensional.

This is equivalent to proving that the closed unit ball in $(S,\|\|_\infty)$ or in $(S,\|\|_2)$ is compact, but I can't derive this. 
Another thought is that the $L^2$ closure of $S$ is a subset of $C([0,1])$. Indeed, if $f_n\to f\in L^2$, then the given relation $\|f_n-f\|_\infty \leq c\|f_n-f\|_2$ implies that $f_n\to f$ in $L^\infty$, so $f$ is continuous. Therefore we have the inclusion
$$S\subset \overline{S}^{L^2}\subset C([0,1])\subset L^2$$
If $S$ was infinite dimensional, then $\overline{S}$ would be an infinite dimensional hilbert space, proper subset of $L^2$. 
Another thought is that all the $L^p$ norms on $S$ are equivalent (by Holder). If ALL norms are equivalent, then $S$ has to be finite dimensional. 

You can apply a theorem of Grothendieck to the closure of $S$ in $L^2$ which is (as you show) contained in $C([0,1]) \subseteq L^\infty$. 
Grothendieck's theorem says that every closed subspace of $L^p(\mu)$ (where $\mu$ is a probability measure on some measurable space and $0<p<\infty$) which is contained in $L^\infty(\mu)$ is finite dimensional.
A convenient reference is Rudin's Functional Analysis, Theorem 5.2. 
