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Can anyone give an example of a non convergent net/ filter in a co-countable space in which the underlying set is un countable. (Such a net/filter exists since any co-countable space in which underlying set is uncountable is not compact).

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    The only thing you need to use about your space is that intersections of 2 non-empty open sets are non-empty. ​ ​ – 2017-01-30
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    Look at the proof that such a net/filter exists on any noncompact space. It should give you a recipe to find an example starting from any open cover of your space with no finite subcover. Alternatively, just think about sequences and when they will converge. – 2017-01-30
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    You need (negation of compactness) a filter such that no superfilter converges or a net such that no subnet converges, BTW. – 2017-01-30

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Let $X$ be any uncountable set and $(a_n)$ any sequence in it with all distinct elements. It is well-known that this sequence cannot converge to any point of $X$, when $X$ has the cocountable topology. This already defines a non-convergent net. So the tail filter also does not converge:

define $T_n=\{a_k: k \ge n\}$, and $\mathcal{F} =\{A \subseteq X: \exists n: T_n \subseteq A\}$

This defines a filter on $X$, and without a convergent larger filter as well.

Suppose $\mathcal{G} \supseteq \mathcal{F}$ is a filter and suppose $\mathcal{G} \rightarrow p \in X$ in the co-countable topology.

Define $C = \{a_n: n \in \mathbb{M}\} \setminus \{p\}$. Then $p \in O = X \setminus C$ and $O$ is open in the co-countable topology, by definition. Then by the supposed convergence, $O \in \mathcal{G}$. Now $C$ contains some $T_n$ (this will be a non-trivial tail if $p$ is one of the sequence points, otherwise $T_0$). So $C \in \mathcal{F} \subseteq \mathcal{G}$. But $C$ and $O$ are disjoint, contradicting the filter property of $\mathcal{G}$.