$\underset{v \in S_h}{\Vert u - v_h \Vert_{1,\Omega}} \leq ch\Vert u \Vert_{2,\Omega} $ implies $H^2(\Omega) \hookrightarrow H^1(\Omega)$

Question

We have the following that is derived from the Bramble-Hilbert Lemma

$\underset{v \in S_h}{\Vert u - v_h \Vert_{1,\Omega}} \leq ch\Vert u \Vert_{2,\Omega} $ implies $H^2(\Omega) \hookrightarrow H^1(\Omega)$

Where $h> 0$ and $S_h$ is finite dimensional.

I have an outline of what the solution is. However, I cannot (1) further it and I cannot (2) sketch it. I believe that if I can properly sketch the following outline then I will understand the abstract concept more from an engineering perspective. The steps are:

1. Let $B$ be a unit ball in $H^2(\Omega)$
2. Choose $h$ such that $ch < \epsilon$ as a result of $u\in B$
3. Find $v_h$ such that $\Vert u - v_h \Vert_{1,\Omega} \leq \epsilon$
4. $\dim(S_h) $ is finite such that we have a bounded set $\{ v \in S_h;\Vert v \Vert_1 \leq 1 \}$. By the idea of totally-bound space or pre-compact space we have that it can be covered by a finite number of balls of radius $\epsilon$ or diameter $2\epsilon$. This is represented by the first image.
5. Now here comes the part that I don't understand. If the diameter of these balls are doubled then they cover the set $B$. Hence $B$ is pre-compact and the completeness of the Sobolev space implies compactness. How does this last bit relate to $H^2(\Omega) \hookrightarrow H^1(\Omega)$ ?

$\textbf{Note:}$ $I_h$ is the interpolating polynomial in the skecth. I believe that $I_h u = u$ as a result that maps $u$ from $H^2(\Omega)$ to $H^1(\Omega)$

I have sketched an attempt. It may be completely wrong. If so, please point out where the fatal mistakes occur.

Thank you so much for your time. I am trying to understand the theory behind finite elements but for someone from an engineering background this is proving difficult.