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 $\mathcal{D}(\Omega)$.
The answer to the central question is: "no".
No matter how you interpret "limit point" it should imply "in the closure of".
If $C=\{a_n:n\in\mathbb{N}\}$ is a sequence, a countable set, in $A$ then $C$ is closed in the co-countable topology and if $p\notin A$ then $\mathbb{R}\setminus C$ is a neighbourhood of $p$ that is disjoint from $C$.
So even if $p$ outside $A$ is in the closure of $A$ it is not in the closure of any countable subset of $A$.
Best Answer
Let $X = \omega_1\times [0, 1) \cup \{\omega_1\}$ where $\omega_1\times [0, 1)$ is given lexicographic order and $\omega_1$ is a point which is greater than all points of $\omega_1\times [0, 1)$.
Give $X$ the order topology.
Then $X$ is connected and $\omega_1$ is not a limit of a not eventually constant sequence.
Imagine $X$ as a long closed interval, modification of the long ray to which we add a top element $\omega_1$, or as the ordinal $\omega_1+1$ to which we fill in the gaps between ordinals using segmemts.