We can concluded that $\mathcal{D}(\Omega):=\bigcup_{K \in \mathcal{K}(\Omega)} \mathcal{D}_K(\Omega)$ (where $\mathcal{K}(\Omega)$ denotes the union of all compacts set content in a open subset $\Omega \subset \mathbb{R}^n$) it's a countable union of Fréchet spaces, and $\mathcal{D}(\Omega)$ is a locally convex space not metrizable. We can also conclude that space of distributions $\mathcal{D}'(\Omega)$ is a locally convex space.

More specifically, for the second question, it's correct to say that $(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ is a dual pair, since $\langle \varphi , u \rangle := u(\varphi)=0$ $\forall u \in \mathcal{D}'(\Omega)$ implies $\varphi=0$ in the sense that since $\mathcal{D}(\Omega) \subset L^1_{loc}(\Omega) \subset \mathcal{D}'(\Omega)$ we have $\int_{\Omega} |\varphi|^2 dx = \langle \varphi , \overline{\varphi} \rangle =0$. Therefore, $\mathcal{D}'(\Omega)$ is equipped with the weak* topology $\sigma(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ making it a locally convex space, with topology defined by sufficient (separable) family of seminorm $\mathcal{F}:=\lbrace p_{\varphi}(u):=|u(\varphi)| : \varphi \in \mathcal{D}(\Omega) \rbrace $ which it is the restriction ad $\mathcal{D}'(\Omega) \subset \mathbb{K}^{\mathcal{D}(\Omega)}:=\prod_{x \in \mathcal{D}(\Omega)} \mathbb{K}$ of the product topology or the topology of pointwise convergence. In particular, it is the minimal topology that makes continuous distributions: okay, $\mathcal{D}'(\Omega)$ is a locally convex space, it's correct?

Therefore we say that $u_k \rightarrow u$ if $\langle \varphi , u_k \rangle \rightarrow \langle \varphi , u \rangle$ $\forall \varphi \in \mathcal{D}(\Omega)$. Many authors, call this property "sequential continuity" of a distributions sequence $\lbrace u_k \rbrace \subset \mathcal{D}'(\Omega)$, but it would be more appropriate to call it weak* convergence. I do not know if you agree.

I'm getting another doubt, that in the space of Fréchet $\mathcal{D}_K(\Omega)$ is defined topology $\mathcal{T}_K$, through seminorm

$\displaystyle p_N(\varphi)=\sup_{|\alpha| \leq N} \left \| D^\alpha \varphi \right \|_{K_{N}}$ , $\displaystyle \left \| D^\alpha \varphi \right \|_{K_{N}}:=\sup_{x \in K} |D^\alpha \varphi|$

and $\varphi_k \rightarrow \varphi$ in $\mathcal{D}(\Omega)$ if and only if

(1) $\exists K \in \mathcal{K}(\Omega): \mathrm{supp}(\varphi_k), \mathrm{supp}(\varphi) \subset K$ $\forall k \in \mathbb{N}$.

(2) $D^\alpha \varphi_k \rightarrow D^\alpha \varphi$ uniformly on $K$ $\forall \alpha \in \mathbb{N}^n$

By definition, a distribution is a continuous linear functional than the previous convergence. In particular, this convergence has compared a locally convex topology $\mathcal{T}$ on $\mathcal{D}(\Omega)$, where it's local base $\mathcal{U}$ is formed by the family of all convex balanced subsets $U$ such that $U \cap \mathcal{D}_K(\Omega)$ is an open subset in $\mathcal{D}_K(\Omega)$.

Now, on $\mathcal{D}(\Omega)$, there is also the weak topology $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega))$ defined by sufficient (separable) family of seminorm $\mathcal{F}=\lbrace p_u(\varphi)=|u(\varphi)| : u \in \mathcal{D}'(\Omega)\rbrace$ and $\varphi_k \rightharpoonup \varphi$ if $u(\varphi_k) \rightarrow u(\varphi)$ $\forall u \in \mathcal{D}'(\Omega)$.

then we can say that $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega)) \subset \mathcal{T}$ ?