Let $\Omega\subset\mathbb R^n$ be an open set, and recall $L^2$ and $H^1$ are the Hilbert spaces with inner products $(u,v)_{L^2}=\int_\Omega u\overline v\,dx$, and $(u,v)_{H^1}=\int_\Omega u\overline v+Du\cdot D\overline v\,dx$.
Fake Claim: for all $u$ in $H^1$, there is a $C>0$ such that $$ \|u\|_{H^1}\le C\|u\|_{L^2}. $$
This is absurd, of course, given the example $u=e^{it\cdot x}$ for $|t|\to\infty$. But the compactness proof below indicates otherwise. Where does it go wrong?
Fake proof: If not, there is a sequence $u_k\in H^1$ such that $\|u_k\|_{H^1}=1$ but $\|u_k\|_{L^2}\to 0$ as $k\to\infty$. By the Banach-Alaoglu theorem, we can find $u\in H^1$ and a subsequence, also denoted $u_k$, such that $u_k\to u$ weakly in $H^1$, i.e. $(u_k,\varphi)_{H^1}\to (u,\varphi)_{H^1}$ for all $\varphi\in H^1$. Note that $\|u\|_{H^1}=1$.
Given this, let now $\varphi\in C^\infty_0(\Omega)$, and consider $(u_k,\varphi)_{H^1}$. Since $Du_k\in L^2$ is the weak derivative of $u_k$, we have $(Du_k,D\varphi)_{L^2}=-(u_k,\Delta\varphi)_{L^2}$ (i.e. integration by parts), where $\Delta=\nabla\cdot\nabla$ is the Laplacian. Therefore,
$$ (u_k,\varphi)_{H^1}=(u_k,\varphi)_{L^2}-(u_k,\Delta\varphi)_{L^2}\to 0 $$ as $k\to\infty$. This would seem to imply $(u,\varphi)=0$ for all $\varphi\in C^\infty_0$, which then implies $u=0$, a contradiction to $\|u\|_{H^1}=1$.
Here is at least one mistake in the above. The unit sphere is not closed in the weak topology on any infinite dimensional normed space. In fact, for any such space $X$, the weak closure of $S = \{x \in X: \|x\| = 1\}$ is the unit ball $B = \{x \in X: \|x\| \leq 1\}$.
In particular, you cannot conclude that $\|u\|_{H^1} = 1$. The rest of the proof (if I haven't missed another mistake) can then be seen as a proof that, in fact, $u = 0$.