Let $\Omega \subseteq \mathbb{R}^n$ be a nonempty open set, and $\mathcal{D}(\Omega)$ the
space of test functions (that is infinitely differentiable functions
$f:\Omega \rightarrow \mathbb{C}$ with compact support contained in $\Omega$), with the usual
topology defined through inductive limit of Fréchet spaces (see Distribution or any good book which deals with distributions as Rudin, Functional Analysis, or Reed and Simon, Methods of Mathematical Physics, Volume I or the wonderful Schwartz, Théorie des Distributions).
Now, let us recall that a topological space $X$ is called a Fréchet-Urysohn space if for every $A \subseteq X$, the closure of $A$ coincides with the sequential closure $[A]_{seq}$ of $A$, which is defined as
\begin{equation}
[A]_{seq} = \{ x \in X : \exists (a_j)_{j=0}^{\infty} : a_j \rightarrow x \textrm{ and } a_j \in A \textrm{ for } j=0,1,2,\dots \}.
\end{equation}
A set $A \subseteq X$ for which $A= [A]_{seq}$ is called sequentially closed. Note that every closed set $A$ is sequentially closed. If also the converse is true, that is if every sequentially closed set turns out to be closed, the space $X$ is called a sequential space (see Sequential Space for more details and references about sequential spaces).
Clearly, every Fréchet-Urysohn space is a sequential space, but the converse is not true (see the post Understanding two similar definitions).
With this terminology we may ask: is $\mathcal{D}(\Omega)$ a Fréchet-Urysohn space? Is it a sequential space?
The answer to these two questions is negative, as I will show in my answer below. I posted the question here only to share this result with the community of math.stackexchange.com since I coud not find the answer in any book I consulted.
Best Answer
I will make use here of the notation introduced in my previous post Topology of the space $\mathcal{D}(\Omega)$ of test functions. We shall show that $\mathcal{D}(\Omega)$ is not a sequential space, so that the answer to both questions is negative, as announced.
Take $V$ as in my answer to the post Topology of the space $\mathcal{D}(\Omega)$ of test functions and let $A$ be the complement of $V$. Then the argument given in that answer shows that $0$ is a limit point of $A$, so $A$ is not closed. Anyhow, $A$ is sequentially closed, as we shall now show.
Suppose that $f \in V$ and that $(f_j)$ is a sequence in $\mathcal{D}(\Omega)$ converging to $f$. Then by the characterization of converging sequences in $\mathcal{D}(\Omega)$ (see e.g. Theorem (6.5) in Rudin, Functional Analysis, 2nd Edition), we know that:
(i) there is a compact set $K$ contained in $\Omega$, such that the support of $f_j$ is contained in $K$ for all $j=0,1,2,\dots$,
(ii) for every $\epsilon > 0$ and every nonnegative interger $N$ there exists a nonnegative integer $m$ such that $\left| \left| f_j - f \right| \right|_N < \epsilon$ for all $j \geq m$.
Now, since $V \cap \mathcal{D}_K \in \tau_k$, there exists $\epsilon > 0$ and a nonnegative integer $N$ such that the set \begin{equation} B = \{ g \in \mathcal{D}_K : \left| \left| g - f \right| \right|_N < \epsilon \} \end{equation} is contained in $V \cap \mathcal{D}_K$. Then, if $m$ is the nonnegatve integer whose existence is stated in (ii), we conclude that $f_j \in V$ for all $j \geq m$. So there is no sequence $(f_j)$ in $A$ converging to $f$.
QED
NOTE. From NOTE (2) in my answer to the post Topology of the space $\mathcal{D}(\Omega)$ of test functions, we know that $A$ is dense in $\mathcal{D}(\Omega)$, and since $A$ is sequentially closed, we can conclude that $A$ is an example of a dense subset of $\mathcal{D}(\Omega)$ which is not sequentially dense in $\mathcal{D}(\Omega)$.