Size of the closure of a set 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.
 A: 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.
A: The statement is consistently false.
In vaughn's article "Countably compact and sequentially compact spaces" in the Handbook of Set-Theoretic Topology, he gives the following equivalences for an infinite cardinal $\kappa$:


*

*$\{0,1\}^\kappa$ is sequentially compact

*$\kappa < \mathfrak{s}$ (where $\mathfrak{s}$ is the "splitting number")

*every compact space of weight $\leq\kappa$ is sequentially compact


Thus if we are in a universe where $\omega<\omega_1=\kappa<\mathfrak{s}\leq\mathfrak{c}<2^\kappa$ then the space
$\{0,1\}^{\omega_1}$ gives a separable space which is sequentially compact but has cardinality greater than $\mathfrak{c}$.
Added: including the forcing construction to get such a model (from van Douwen's article):
Start with ground model
$M\models(\mathfrak{c}=\omega_2 \wedge 2^{\omega_1}=\omega_3)$
then obtain an iterated ccc extension $\langle M_\eta : \eta\in\omega_2\rangle$ by adding $X_\eta\in[\omega]^\omega$ at stage $\eta\in\omega_2$ s.t. $\forall Y\in M\cap[\omega]^\omega$ either $X_\eta\subseteq^* Y$ or $X_\eta\subseteq^* (\omega\setminus Y)$.  Then in $M_{\omega_2}$ we have $\mathfrak{c}=\omega_2<2^{\omega_1}$, $\mathfrak{t}=\omega_1$, $\mathfrak{s}=\omega_2$, exactly as desired.  (Details are left as an exercise).
A: It seems like you don't need all the conditions, so maybe I am missing something.  There are only continuum many sequences of points from a countable set.  So the limit points are at most continuum many.
