Does every point of the closure of a set have a sequence that converges to itself?

Question

I'm working on the following problem:

Extension of a Uniformly Continuous Function between Metric Spaces

But I'm terribly confused about the relationship between limit points and closure of a set. In the second answer in this post above, they say

If $a\in \overline{A}$, then $a=\lim_n a_n$, where $a_n\in A$.

Why is this true? I think I must've forgotten something very fundamental here, however, as far as I can tell, this is true if this $a$ is a limit point of $A$. But this $a$ is taken from $\overline{A}$, not $A'$, i.e. the set of all limit points of $A$.

Answer 1

We have $\overline{A} = A\cup A'$. If $a\in A'$, the answer is clear. If $a\in A$, then simply we can take $a_n = a$ for all $n$.

But if we give the condition $a_n\neq a$, then the statement is not true in general. If $A$ has an isolated point(i.e. $a\in A\setminus A'$), then there's no such sequence. For example let $A = [-1, 1]\cup \{2\}\subset\mathbb{R}$ and $2$ is an isolated point of $A$.