Consider the following operation on a subset $A$ of a space $X$, defining a new subset of $X$: $$\mbox{s-cl}(A) = \{ x \in X \mid \mbox{ there exists a sequence } (x_n)_n \mbox{ from } A \mbox{ such that } x_n \rightarrow x \}\mbox{.}$$ This set, the sequential closure of $A$, contains $A$ (take constant sequences) and in all spaces $X$ it will be a subset of the $\mbox{cl}(A)$, the closure of $A$ in $X$.
We can define $\mbox{s-cl}^{0}(A) = A$ and for ordinals $\alpha > 0$ we define $\mbox{s-cl}^\alpha(A) = \mbox{s-cl}(\cup_{\beta < \alpha} \mbox{s-cl}^\beta(A))$, the so-called iterated sequential closure.
A space is Fréchet-Urysohn when $\mbox{s-cl}(A) = \mbox{cl}(A)$ for all subsets $A$ of $X$, so the first iteration of the sequential closure is the closure.
A space is sequential if some iteration $\mbox{s-cl}^\alpha(A)$ equals the $\mbox{cl}(A)$, for all subsets $A$.
So basically by taking sequence limits we can reach all points of the closure eventually in a sequential space, but in a Fréchet-Urysohn space we are done after one step already.
For more on the differences and the "canonical" example of a sequential non-Fréchet-Urysohn space (the Arens space), see this nice topology blog, and the links therein.
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 $\mathcal{D}(\Omega)$.
Best Answer
It boils down to the following. (Note: no Hausdorff assumption needed.)
Suppose $X$ is a countable topological space and $A\subseteq X$. If there is a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points of $A$ that converges to a point $p\in X$, then there is an ordinary sequence ($\omega$-indexed) of points of $A$ converging to $p$. Here $\lambda$ is some limit ordinal and saying the transfinite sequence converges to $p$ means: for every nbhd $V$ of $p$, there is some $\beta<\lambda$ such that $x_\alpha\in V$ for all $\alpha>\beta$.
By considering the cofinality of $\lambda$ and taking a corresponding suitable cofinal subsequence of $(x_\alpha)_{\alpha<\lambda}$, we can assume $\lambda$ is a regular cardinal and then we can further assume it is of smallest (infinite) cardinality satisfying the conditions.
Now, if one of the $x_\alpha=a\in A$ is repeated cofinally many times, then extracting the subsequence of all the elements with value $a$ gives a constant (transfinite) sequence that converges to $p$. (Note that $a$ need not be equal to $p$.) Then, by definition of convergence, every nbhd of $p$ contains $a$ and the ordinary sequence $(y_n)_{n<\omega}$ with each $y_n=a$ also converges to $p$.
Otherwise, for each $a\in A$ the set of indices $I_a=\{\alpha<\lambda: x_\alpha = a\}$ is not cofinal in $\lambda$, that is, $I_a$ has cardinality smaller than $\lambda$. But there are only countably many values $a\in A$. So we get a partition of $\lambda$ into countably many sets, each of cardinality smaller than $\lambda$. If $\lambda$ were uncountable, that would contradict the fact that $\lambda$ is a regular cardinal. So necessarily $\lambda=\omega$ and our original sequence was an ordinary sequence.
This proof also applies to a locally countable space, by first restricting the sequence to a tail contained in a countable nbhd of $p$.
[Added later] From the proof above, all that matters is that $A$ is countable, even if $X$ is not. So basically the result is:
An attempt at a more handwaving proof would be: (1) if an element is repeated cofinally many times, do the same as the above. (2) otherwise, for each repeated element in the sequence, take its first appearance and remove all the later copies. The resulting subsequence should also converge to $p$. Why? Because it's cofinal in the original sequence. And why is that? It's not completely obvious, but the argument above justifies it rigorously.
[As mentioned in Steven's comment below, this handwaving attempt does not work.]
(added 11/15/2023)
For more general cardinalities one can use essentially the same argument as above to show for any topological space $X$:
Proposition: Suppose the point $p\in X$ is the limit of a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points from a set $A\subseteq X$ of infinite cardinality $\kappa$. (Here $\lambda$ is some limit ordinal.) Then $p$ is also the limit of a transfinite sequence of points of $A$ indexed by some infinite regular cardinal $\mu$ with $\mu\le\min(\kappa,\lambda)$.
Note: This applies in particular when $A$ is the set of values of the transfinite sequence. Also, if $\mu$ is finite, there is a constant (ordinary) sequence with value in $A$ converging to $p$.