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 $\|\cdot\|_{\infty}$.