Intersection of nested closed sets in Hausdorf topological space I am trying to prove the following statement, but starting to doubt its correctness.
Suppose that $H$ is a Hausdorf topological space (I am formulating generally, though my specific case is $H=S'(R)$ - a space of tempered distributions). 
Suppose I have a set of nested subsets $\Omega_i \subseteq H$ and $\Omega_i \supseteq \Omega_{i+1}$ and by $\overline{\Omega}$ we denote a sequential closure of $\Omega\subseteq H$. Is it true that:
$x\in \cap_{i=1}^\infty \overline{\Omega_i}$ if and only if there is sequence $\{x_i\}$ such that $x_i \rightarrow x, x_i\in \Omega_i$. 
The fact that from $x_i \rightarrow x, x_i\in \Omega_i$ we can deduce $x\in \cap_{i=1}^\infty \overline{\Omega_i}$ is obvious. The opposite is problematic.
I tried to prove the opposite statement via reducing to Cantor's intersection theorem. Suppose that $x\in \cap_{i=1}^\infty \overline{\Omega_i}$ is fixed. Then I define 
$R_i = \{\{x_n\}| \exists N, \forall n>N: x_n \in \Omega_i, x_n \rightarrow x\}$ (a set of sequences that tend to $x$ and is in $\Omega_i$ starting from some index). It is easy to see that $R_i$ is also nested: $R_i\supseteq R_{i+1}$.
Then the wanted statement is equivalent to $\cap_{i} R_i \ne \emptyset$. 
The problem now is that I need compactness of $R_i$ in order to apply Kantor's theorem, but I stuck at this step.
 A: For a counterexample using sequential closures, consider the topological space whose underlying set is $\mathbb{N}^2 \sqcup \{ x_0 \}$, and with the topology such that $U$ is open if and only if $x_0 \notin U$ or for some function $f : \mathbb{N} \to \mathbb{N}$, $\{ (x, y) \in \mathbb{N}^2 \mid y > f(x) \} \subseteq U$.
Now, let $\Omega_n := \{ (x, y) \in \mathbb{N}^2 \mid x \ge n \}$.  Then $x_0$ is in the sequential closure of $\Omega_n$ for each $n$ since $(n, m) \to x_0$ as $m \to \infty$.  On the other hand, if we have any sequence $(x_n, y_n) \in \Omega_n$, then $x_n \to \infty$ as $n \to \infty$.  Using this, it is possible to construct a function $f : \mathbb{N} \to \mathbb{N}$ such that $y_n < f(x_n)$ for each $n$.  It follows that $(x_n, y_n) \not\to x_0$ as $n \to \infty$ since the corresponding neighborhood of $x_0$ for this $f$ does not contain any element of the sequence.
A: The space of distributions is not even a sequential space (indeed not first countable) so what you are trying to prove has little hope of beeing true. 
If $x \in \cap_i \bar{\Omega_i}$, then for each $i$, there is a sequence $(y_{i,n})_n$ in $\Omega_i$ that converges to $x$, since $x \in \bar{\Omega_i}$. Then what about the following sequence :
$$
x_j = y_{j,j} \; \forall j ? 
$$
Well for each $j$, $x_j = y_{j,j}$ is in $\Omega_j$ by construction.
What about convergence ? Here maybe you need more than just Hausdorff and sequential, lets use metric !
Instead of using the diagonal sequence, build the sequence by induction :
$$
x_0 = y_{0,0}
$$
and
$$
x_{j+1} = y_{j+1,i_j}
$$
where $i_j$ is the smallest integer such that $d(y_{j+1,i_j},x)<2^{-j}$.
I read there was use of first countable neighborhood basis in the comments, it is the same, you need to chose the next candidate of your sequence from $(y_{k,l})$ in such a manner that you get 'closer' to $x$ each time.
In the space of tempered distributions with classical closure definition this statement is false since it is not sequential : 
Suppose the statement is true, consider a familly $\Omega_i$ that is constant, thent that statement for constant famillies of subspaces implies that the space is sequential therefore a contradiction.
