TITLE: Size of the closure of a set
QUESTION [23 upvotes]: Why in a Hausdorff sequentially compact space the size of the closure of a countable subset is less or equal than $c$ ? I can see why this is true when the space if first countable but we are not assuming so.

REPLY [4 votes]: Here is some progress in the positive direction.
Theorem. In any Hausdorff space, the closure of a
countable set has size at most $2^{\mathfrak{c}}$, where $\mathfrak{c}$ is the
continuum.
Proof. Suppose that $X$ is a Hausdorff topological space
with a countable set $D$. For any point $a$ in the closure
$\bar D$, we may consider the collection of open sets
containing $a$, and their trace on $D$. That is, consider
$F_a=\{U\cap D\mid a\in U\text{ open }\}$. Since there are
only continuum many subsets of $D$, there are therefore at
most $2^{\mathfrak{c}}$ many possible such families.
If the closure $\bar D$ had size larger than $2^{\mathfrak{c}}$, then
there would be at least two (in fact many) distinct points
$a$ and $b$ in $\bar D$ for which $F_a=F_b$. Let $U$ and
$V$ be disjoint neighborhoods of $a$ and $b$. Let $U_1$ be
another neighborhood of $a$ such that $U_1\cap D=V\cap D$,
which must exist since $F_a=F_b$. Thus, $U\cap U_1$ is a
neighborhood of $a$ that is disjoint from $D$,
contradicting $a\in\bar D$. QED
The bound is sharp, even for compact Hausdorff spaces, in
the sense that the Stone–Čech 
compactification
$\beta\mathbb{N}$ is a Hausdorff topological space of size
$2^c$ with a countable dense set. But $\beta\mathbb{N}$ is
not sequentially compact, so this is not actually a
counterexample to your question.
What the argument actually shows is that if $a$ and $b$ are
in the closure of the countable set $D$, then $F_b$ is not a
subset of $F_a$. If it were, we could find disjoint
neighborhoods $U$ and $V$ of $a$ and $b$, respectively, and
then $V\cap D\in F_b$, and so there is a neighborhood $U_1$
of $a$ with $U_1\cap D=V\cap D$, making $U\cap U_1$ a
neighborhood of $a$ having no points from $D$, a
contradiction. By symmetry, we conclude $F_a$ and $F_b$ are
incomparable with respect to $\subset$ for all $a,b\in\bar
D$.
I suspect that such a line of reasoning could be improved
when there is sequential compactness, perhaps by using a
cardinal characteristic, such as the splitting number.