Let $X$ be the space $(\omega+1)\times(\omega_1+1)-\{(\omega,\omega_1)\}$, putting the transfinite order topology on each coordinate and the product topology on the whole space.
(0,omega_1) (1,omega_1) (2,omega_1) ---> O
: : :
: : : :
: : : :
: : : :
(0,alpha) (1,alpha) (2,alpha) ---> (omega,alpha)
: : : :
: : : :
(0,1) (1,1) (2,1) ---> (omega,1)
(0,0) (1,0) (2,0) ---> (omega,0)
Thus, the space consists of $\omega_1+1$ many copies of a (horizontally) convergent sequence, stacked on top of each other, but with the limit of the final sequence omitted (at O in the diagram). Horizontally, the $\alpha$-th row has $(n,\alpha)$ converging to $(\omega,\alpha)$ as $n\to\omega$. Vertically, the $n$-th column has a transfinite sequence $(n,\alpha)$ converging to $(n,\omega_1)$ as $\alpha\to\omega_1$. This is a Hausdorff space, and comparatively nice in several ways.
Let $A=(\omega+1)\times\omega_1\subset X$ be the lower portion of this diagram, without the top row. Any countable sequence from $A$ is bounded below a countable ordinal and has accumulation points in $X$, and in fact in $A$ itself: either the sequence contains infinitely many points from one coordinate, in which case it accumulates on the limit supremum in that coordinate, or it contains points from infinitely many coordinates, which will accumulate on a point of the form $(\omega,\alpha)$ at the right-most column in $X$. In particular, $A$ already contains all its limit points of sequences in $A$. Nevertheless, the closure of $A$ in $X$ is all of $X$, and this includes the top row of points of the form $(n,\omega_1)$. But this sequence has no accumulation points in $X$, as desired.