First of all, I know a similar question has been asked here compactness in topology of pointwise convergence,
but I am still do not know how to identify compact subsets. Given a set $X$, endowed with the topology of pointwise convergence, such that it inherits the product topology from $\mathbb{R}^{\mathbb{N}}$. If a subset $U \subset X$ is bounded and closed, is it automatically compact? Does $X$ "inherit" the Heine-Borel-Theorem?
It's certainly not true in the usual metric on $\mathbb{R}^\mathbb{N}$, which is defined by
$$d((x_n)_n, (y_n)_n) = \sum_{n=0}^\infty \frac{\min(\left|x_n - y_n\right|,1)}{2^n}$$
where we need the truncated standard metric to ensure convergence of the sum. This $d$ induces the pointwise (product) topology on $\mathbb{R}^\mathbb{N}$ and all subspaces. But the whole metric is bounded by $2$, so Heine-Borel won't hold.
I strongly doubt that you could find a metric on this space that induces the product topology and satisfies the Heine Borel theorem.