Equivalent affirmation of "$\{a_n\}$ has no convergent subsequences" I'm trying to prove that a closed set $K$ in a normed vector space is compact if

For every family $\{F_n\}$ of nested non-empty closed subsets of $K$, $\bigcap_{n \ge 1} F_n \ne \emptyset$

I'm trying to prove this with sequences, I think that if I manage to prove that if $\{a_n\}$ has no convergent sequences, then it's closed. I'll be able then to apply the propierty to the family of subsequences $\{a_k\}_{k \ge n}$ since they are nested, and since their intersection is non-empty, then $\{a_n\}$ must be convergent (a contradiction). Given this contradiction, then $\{a_n\}$ has a convergent subsequence, thus $K$ is compact.
I have a feeling that I'm asking too much in this pseudo proof, since I don't really have a solid expression to the proposition "$\{a_n\}$ has no convergent subsequence"
 A: (1). It is not the case that in every topological space, that the closure operation, or the property of being a closed set, can be defined in terms of convergent sequences.
$$ \text {Definition: }\quad   \lim_{n\to \infty}b_n=b  \iff  \{b\}=\cap_{n\in \mathbb N}Cl(\{a_m:m\geq n\}).$$
Consider this example: $X=\mathbb Z \cup \{-\infty,\infty\}$ with $\mathbb Z \cap \{-\infty,\infty\}=\phi$ and $-\infty \ne \infty.$ For $Y\subset X$ let $Y$ be open iff (a): $Y\subset \mathbb Z,$ or (b): $\{-\infty,\infty\}\subset Y$ and $\mathbb Z$ \ $Y$ is finite. The sequence $(n)_{n\in \mathbb N}$ has no convergent subsequences but $\{a_n:n\in \mathbb N\}=\mathbb N$ is not closed in $X.$  However $X$ is compact.
There are also examples where $X$ is a compact Hausdorff space in which closure is not definable in term of convergent sequences.
And there are examples where $X$ is a locally compact, non-compact Hausdorff space in which every sequence has a convergent sub-sequence, and closure IS definable in terms of convergent sequences. E.g., the $\epsilon$-order topology on $\omega_1$.
(2). Since closed sets are the complements of open sets, any assertion about open sets (e.g. compactness) has a corresponding, equivalent,  dual  assertion about closed sets.
Here is a way to solve it without sequences:
Suppose $F$ is a non-empty family of non-empty closed sets of the space $X,$ such that (a):$\forall f,g\in F\;(f\subset g\lor g\subset f),$ and (b): $\cap F=\phi.$
Let $G$ be the set of finite  non-empty subsets of $F$. Any $H\in G$ has a subset-minimum member $h,$ so $\cap H=h.$ And $h\ne \phi$ because $h\in F. $
Observe that $\cap_{H\in G}(\;\cap H\;)=\cap F=\phi.\quad$ So  $J=\{X \backslash \cap H:H\in G\}$ is an open cover of $X.$ 
But $J$ has no finite sub-cover. Because if $\{H_1,...,H_n\}\subset G,$ then $H=\cup_{i=1}^nH_i\in H$ and $$\cap_{i=1}^n (\;\cap H_i\;)=\cap H\ne  \phi,$$ $$ \text {so }\quad \cup_{i=1}^n(X \backslash \cap H_i)=X \backslash \cap H\ne X. $$
BTW the precise dual of "Every open cover of $X$ has a finite sub-cover" is (equivalent to): "If $F$ is  a non-empty closed family with the F.I.P. then $\cap F \ne \phi\;$"..... where a closed family means a family of closed sets, and F.I.P. (Finite Intersection Property) means that $\cap H\ne \phi$ whenever $H$ is finite and $\phi \ne H\subset F.$
