On continuity and topology in the kernel theorem of Schwartz

I am aware that this might be a basic question, but I am relly get messed up regarding various different topologies in functional analysis discussed in distirbution theory.

The well-known kernel theorem of Schwartz assersts basically the following: Let $X\subset\mathbb{R}^{d}$ be open. Then, a linear operator $A:C^{\infty}_{c}(X)\to\mathcal{D}^{\prime}(X)$ is continuous if and only if there exists a bidistribution $k_{A}\in \mathcal{D}^{\prime}(X\times X)$ such that $$\langle A\varphi,\psi\rangle_{X}=\langle k_{A},\psi\boxtimes\varphi\rangle_{X\times X}$$

In the formulation above, which topology is chosen on $\mathcal{D}^{\prime}$? Is it the strong one or the weak $\ast$-one? Furthermore, the continuity of $A$, is it supposed to be sequential continuity only? I am asking this because I have seen different claims in different sources. In most claims, it seems that one takes the strong topology on $\mathcal{D}^{\prime}$ and strong continuity for $A$. However, in the German wikipedia page, it is somehow stressed that an operator $A$ coming from a kernel $k_{A}$ is in general only sequentially continuous, but they do not specify which topology they use on $\mathcal{D}^{\prime}$. Last but not least, in this book of Tarkhanov (see section 1.5.1), the operator $A$ is in the space $\mathcal{L}_{b}(C^{\infty}_{c}(X),\mathcal{D}^{\prime}(X))$, which is the topology of uniform convergence on bounded subsets of $C^{\infty}_{c}(X)$. How does this fit into the picture?

The space $\mathscr D=C_c^\infty(X)$ of test functions gets the locally convex inductive limit topology (=colimit in the locally convex category), which is the finest locally convex topology on $\mathscr D$ such that all inclusions $i_K$ from $\mathscr D_K =\{\varphi\in\mathscr D:$ supp$(\varphi)\subseteq K\}$ into $\mathscr D$ for compact subsets $K\subseteq X$ are continuous. The spaces $\mathscr D_K$ are Fréchet spaces in the natural topology of uniform convergence of all (partial) derivatives. The universal property of colimits implies that a linear $A:\mathscr D\to F$ into any locally convex space is continuous if and only if all compositions $A\circ i_K$ (the restrictions of $A$ to $\mathscr D_K$) are continuous.

Since $\mathscr D_K$ are metrizable you get that continuity of $A$ is equivalent to sequential continuity. Moreover, if the range $F$ is a so-called webbed locally convex space -- a class introduced by de Wilde in response to a question of Grothendieck about generalizations of the closed graph theorem, in particular $F=\mathscr D'(X)$ with its strong topology is webbed -- then you have a closed graph theorem: If $A:\mathscr D \to F$ has (sequentially) closed graph then $A$ is continuous.

In particular, if $A:\mathscr D \to (\mathscr D'(X),\sigma(\mathscr D'(X),\mathscr D))$ is sequentially continuous then it has closed graph and hence it is continuous as an operator $\mathscr D\to (\mathscr D'(X),\beta(\mathscr D'(X),\mathscr D))$.

The space of test functions is a Montel space, and on its continuous dual strong convergence and weak-* convergence are the same for sequences. So it doesn't matter which of the two you pick in regards to sequential continuity.

In

Trèves, François, Topological vector spaces, distributions and kernels, Mineola, NY: Dover Publications (ISBN 0-486-45352-9). xvi, 565 p. (2006). ZBL1111.46001.

these statements are proven in Chapter 34, section 4.