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.
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
$\kappa &lt; \mathfrak{s}$ (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 $\omega&lt;\omega_1=\kappa&lt;\mathfrak{s}\leq\mathfrak{c}&lt;2^\kappa$ 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 $\mathfrak{c}=\omega_2&lt;2^{\omega_1}$, $\mathfrak{t}=\omega_1$, $\mathfrak{s}=\omega_2$, exactly as desired.  (Details are left as an exercise).
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.