Does Arzelà-Ascoli theorem hold for pointwise convergence? Consider the space $C(\left[0,1\right])$ of continuous, real-valued functions on the interval, equipped with the topology of pointwise convergence.

Is it true that a subset $S\subseteq C(\left[0,1\right])$ is compact if and only if it is bounded and equicontinuous? 

I guess that the answer is no, as this seems to be a weakened version of Arzelà-Ascoli theorem, which guarantees the validity of the statement when the topology is given by uniform convergence instead. I was trying to find a counterexample but it's not easy to check that a certain subset is or is not compact, especially because pointwise convergence topology is not metrisable, so we cannot use sequential compactness arguments. Can you help me with that? Thank you.
As a starting point we need to take the correct version of Arzela-Ascoli for the compact-open (or in this case also uniform) topology: $S$ is relatively compact (the closure is compact) iff it is pointwise relatively compact, and equicontinuous.

In the weaker pointwise topology, we have more compact sets, and there the criterion is simply: $S$ is relatively compact iff it is pointwise relatively compact: this makes it a subset of a product space of compact intervals of the reals, and thus has a compact closure due to the Tychonoff theorem.

In both cases, to characterize compactness, add closed to the list of conditions.
It might be helpful if you think of C([0,1]) as a subspace of the (compact,hausdorff) product space $R^{[0,1]}$. 
- Apologies, this is not correct!