If $\{h_n\}_{n∈N} ⊂ C ([a, b])$ is $\| \cdot \|_{\infty}$ convergent to $h$, then $A ={h_n}∪{h}$ is $\| \cdot \|_∞$-compact, $\|\cdot \|_∞$-closed, $\| \cdot \|_∞$-bounded and uniformly equicontinuous.
Is it proved in the same way that I prove $A$ is compact because of the Arzela-Ascoli theorem?
No need for Arzela-Ascoli. In any metric space $x_n \to x$ implies $\{x,x_1,x_2,\cdots\}$ is compact and this follows easily from definition of compactness (using open covers). Just note that $x$ belongs to one of the sets in the open cover and this set also contains $x_n$ whenever $x$ is sufficently large. It follows that there is a finite subcover.
