When is an accumulation point not the limit of some sequence in a topological space?
Here are two examples:
The closed ordinal space $\omega_1 + 1 = [ 0 , \omega_1 ]$ consisting of all countable ordinals as well as the least uncountable ordinal $\omega_1$. It is easy to see that $\omega_1 \in \overline{[ 0 , \omega_1 )}$, but since any sequence in $[0 , \omega_1 )$ is countable, it has an upper bound $\beta$ which is a countable ordinal, and $( \beta , \omega_1 ]$ is an open neighbourhood of $\omega_1$ disjoint from the sequence.
The Stone–Čech compactification $\beta \mathbb{N}$ of the discrete space $\mathbb{N}$. This space has the property that there are no non-trivial (i.e., not eventually constant) convergent sequences. (For quite a bit of basic information about this space see Dan Ma's Topology Blog.) We have that $\mathbb{N}$ is a dense subset of $\beta \mathbb{N}$, but no sequence in $\mathbb{N}$ can converge outside of $\mathbb{N}$.
(Consider also this previous answer of mine, and possibly also this question and its answers.)
For the interested:
One may be lead to believe that the problem is that sequences are "too short", and that things would change if only we were allowed to use "longer" sequences. This is almost true.
Definition. A topological space $X$ is called radial if for any $A \subseteq X$ and any $x \in \overline{A}$ there is a transfinite sequence $\langle x_\xi \rangle_{\xi < \alpha}$ (where $\alpha$ is an ordinal) in $A$ which converges to $x$. (The definition of convergence in this sense is the natural extension of sequential convergence.)
It is not too difficult to show that $\omega_1 +1$ is radial (in fact, we only need to allow sequences of length $\omega_1$ to capture this property).
However $\beta \mathbb{N}$ is not even radial!
Another example: An uncountable space $X$ with countable complement (co-countable) topology.
Proof: Pick any point $x\in X$. It is the accumulation point of $A=X\setminus \{x\}$, however it is not the limit point of $A$. Since for any countable subset $S$ of $A$, $U=X\setminus S$ is an open set of $x$, which satisfies that $U \cap S=\emptyset$.