Norms on C[0, 1] inducing the same topology as the sup norm This is an old homework problem of mine that I was never able to solve.  The solution may or may not involve the Baire category theorem, which I am terrible at applying.
Let $C[0, 1]$ denote the space of continuous functions $[0, 1] \to \mathbb{R}$.  Suppose $|| \cdot ||$ is a norm on $C[0, 1]$ with respect to which the evaluation functions $f \mapsto f(x), x \in [0, 1]$ are continuous.  Show that the topology induced by $|| \cdot ||$ is the same as the usual topology induced by the sup norm.
It is straightforward to show that any sequence converging uniformly converges with respect to $|| \cdot ||$.  I am stuck on proving the converse; I cannot seem to figure out how to use the assumption that $|| \cdot ||$ is a norm. 
Edit:  Zhen Lin informs me that $|| \cdot ||$ is supposed to be complete.  That should make the problem statement true now!
 A: Here's $\DeclareMathOperator{\ev}{ev}$ the answer if $(C[0,1], \|\cdot\|)$ is assumed to be complete. Consider the family $\mathcal{F} = \{\ev_{x}\}_{x \in [0,1]}$ of continuous linear functionals on $(C[0,1],\|\cdot\|)$. For each $f \in C[0,1]$ we have $\sup_{x \in [0,1]} |\ev_{x}(f)| \leq \|f\|_{\infty}$, so the family $\mathcal{F}$ is pointwise bounded. By the uniform boundedness principle the family is uniformly bounded, that is to say $\sup_{x \in [0,1]} \|\ev_{x}\| \leq M$ for some constant $M$. On the other hand $|f(x)| = |\ev_{x}(f)| \leq \|ev_{x}\|\|f\| \leq M \|f\|$ gives $\|f\|_{\infty} = \sup_{x \in [0,1]} |f(x)| \leq M\|f\|$, so the identity $(C[0,1], \|\cdot\|) \to (C[0,1],\|\cdot\|_{\infty})$ has norm at most $M$. Since both spaces are complete, we may apply the open mapping theorem in order to conclude that its inverse is also continuous. In other words, the norms $\|\cdot\|$ and $\|\cdot\|_{\infty}$ are equivalent.

Edit. Here's an example that shows that completeness of the norm is necessary:
Choose a discontinuous linear functional $\varphi: (C[0,1],\|\cdot\|_{\infty}) \to \mathbb{R}$ and define a norm on $C[0,1]$ by
\[
\|f\| = \|f\|_{\infty} + |\varphi(f)|.
\]
Since $\|f\|_{\infty} \leq \|f\|$, we have that the identity $(C[0,1],\|\cdot\|) \to (C[0,1],\|\cdot\|_{\infty})$ is continuous. Since the evaluation functionals are continuous with respect to the sup-norm, they are also continuous with respect to the norm $\|\cdot\|$. But as $\varphi$ is discontinuous, there is a sequence $f_{n}$ with $\|f_{n}\|_{\infty} = 1$ and $|\varphi(f_{n})| \to \infty$, hence the norms cannot be equivalent. Of course, $\|\cdot\|$ cannot be complete because the last sentence would be in contradiction to the open mapping theorem.

Edit 2.
I forgot to argue why the topologies in the above counterexample are not the same. This is obvious: The functional $\varphi$ is continuous with respect to $\|\cdot\|$ but it isn't continuous with respect to $\|\cdot\|_{\infty}$.
A: I've completely forgotten the solution, but I think something like this might work: The set $\mathcal{F}$ of all evaluation maps is a family of bounded linear operators on $C[0, 1]$. It suffices to show that $\mathcal{F}$ is uniformly bounded, and then the uniform bound gives you the constant needed to establish the equivalence with the uniform norm. I don't remember how to establish uniform boundedness, but I imagine the uniform boundedness principle applies.
