# Equicontinuous, bounded, and closed implies pointwise compact

I'm trying to prove a particular version of the Arzela-Ascoli theorem. I have already gone through the general version:

Let $A \subset M$ where $A$ is compact and $M$ is a metric space.
Let $B \subset C_{b}(A,N)$ where $N$ is a metric space and $C_{b}(A,N)$ is the set of all bounded continuous functions from $A$ to $N$.

Then $B$ is compact if and only if $B$ is equicontinuous, closed, and pointwise compact.

Now I'm trying to prove the specific version. In particular I'm trying to prove the reverse direction of the following:

Let $A \subset M$ where $A$ is compact and $M$ is a metric space.
Let $B \subset C_{b}(A,\mathbb{R}^{m})$ where $\mathbb{R}^{m}$ is a metric space and $C_{b}(A,\mathbb{R}^{m})$ is the set of all bounded continuous functions from $A$ to $\mathbb{R}^{m}$.

Then $B$ is compact if and only if $B$ is equicontinuous, closed, and bounded.

I'm trying to use the general version to prove the specific version. I've managed to show the forward direction (this is trivial using the general version of the theorem). I'm trying to prove the reverse direction.

So I assume equicontinuous, closed and bounded. I will be done if I can show $B$ is pointwise compact, and then apply the general version of the theorem. How can this be done (showing pointwise compactness)?

Notes:
$B$ is pointwise compact if for each $x \in A$, the set $B_{x} = \{ f(x) | f \in B \}$ is compact.