Since each open set $\Omega \subset \mathbb{R}^n$ admits a increasing exhaustion of compact set $K \subset \Omega$, then the space of test functions are rewritten as a countable union $\mathcal{D}(\Omega):=\bigcup_{j \in \mathbb{N}} \mathcal{D}_{K_j}(\Omega)$, and $\mathcal{D}(\Omega)$ is not metrizable, because by contradiction if it is metrizable then it is a complete metric space and by theorem/corollary of Baire $\mathcal{D}_{K_j}(\Omega)$ has interior non empty for some $j \in \mathbb{N}$ and this is a contradiction because $\mathcal{D}_{K_j}(\Omega)$ must necessarily have interior set empty. I justified this because $\mathcal{D}_{K_j}(\Omega)$ is a vectorial topological subspace cloed of $\mathcal{D}(\Omega)$.
Can you have a better formulation of this fact?
The space of distributions $\mathcal{D}'(\Omega)$ is a locally convex space with seminorms family $\mathcal{F}=\lbrace u_{\varphi}(u)=|u(\varphi)| : \varphi \in \mathcal{D}(\Omega) \rbrace$ and $u \in \mathcal{D}'(\Omega)$ which defines the weak* topology $\sigma(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ on $\mathcal{D}'(\Omega)$.
How do I prove that the distribution spaces is not metrizable?
Thanks for any help