Let $U \subset \mathbb{R}^n$. If
$$H^k=\{f: U \rightarrow \mathbb{R}: D^\alpha f \in L^2(U)\ \forall \alpha \in \mathbb{N}^n \ \text{with} \ \vert \alpha \vert \leq k \}$$
then how do I show that
$$\cap _{k \geq 2} H^k \subset C^\infty (U)$$
Is there a dircet approach that does not go through Morrey's and the Gagliardo-Nirenberg-Sobolev inequality?
Yes, but you have to define the Hilbert-Sobolev spaces $H^s(\mathbb{R}^n)$ with $s \in \mathbb{R}$, and this is a corollary to the continuous inclusion theorem Sobolev, which you can view here:
If $u \in H^s(\mathbb{R}^n)$ for $s > n/2$, then $u \in L^\infty(\mathbb{R}^n)$?
and $\bigcap_{s \in \mathbb{R}} H^s(\mathbb{R}^n) \subset C^k(\mathbb{R}^n)$. In particular, if $s=m \in \mathbb{N}$, then $H^s(\mathbb{R}^n)=H^m(\mathbb{R}^n)$ with equivalent norms, and it can be shown that, if $m-k > n/2$, then $H^m(\Omega) \hookrightarrow C^k(\Omega)$, and also $\bigcap_{m \in \mathbb{N}} H^m(\Omega) \subset C^k(\Omega)$, with $\Omega \subset \mathbb{R}^n$ open.