A Banach space $X$ is compactly embedded in $Y$ if $X \subset Y$ and the inclusion $i:X \to Y$ is compact. Since $C([0,1]) \subset L^p([0,1])$, and not the other way around, $L^p([0,1])$ cannot be compactly embedded into $C([0,1])$.
The wikipedia article is exhibiting a compact operator $T:L^p([0,1]) \to C([0,1])$. This does not mean $L^p([0,1])$ is compactly embedded in $C([0,1])$, since this is not the inclusion (and it can't be since $L^p([0,1])$ is much larger than $C([0,1])$).
The Rellich–Kondrachov compactness theorem says that $H^1([0,1])$ is compactly embedded into $L^2([0,1])$. This means the inclusion $i:H^1([0,1]) \to L^2([0,1])$ is compact. In other words, bounded sequences in $H^1([0,1])$ admit subsequences converging in $L^2([0,1])$.
As to the comment in the wikipedia article, the operator $T$ is a bounded linear operator $T:L^2([0,1]) \to H^1([0,1])$. If we compose $T$ with the inclusion $i:H^1([0,1]) \to L^2([0,1])$, then $i \circ T$ is a compact mapping from $L^2([0,1])$ to $L^2([0,1])$, since $i$ is compact. Thus, no need to use Arzela-Ascoli directly here (incidentally, the Arzela-Ascoli theorem is an integral part of the proof of the Rellich-Kondrachov Theorem, so there is no getting away from it!). This is what the wikipedia article is getting at, I believe.
I should note there is nothing special about $p=2$ here. The Rellich-Kondrachov theorem holds for $1 \leq p \leq \infty$, and so the comment above does as well.
A general comment about wikipedia. There are many great math articles and it can be good for reference and learning. However, you should be aware there are many errors, and sometimes explanations are poor or flat out wrong. It is best to look for a good reputable textbook for learning a specific topic in mathematics.