The question I am asking is actually I have asked here.
I have solved the problem in the following way:
Proof: We have every convergent net in $X$ to be eventually constant....$(1)$
We have to show that the Topology with respect to which the nets are convergent on $X$ is Discrete i.e every singleton is open.
Let us define $\mathcal T=\{G\subset X\,:\,\text{ No net in $X\setminus G$ converges in $G$}\}$.
Then obviously $\mathcal T$ is a topology on $X$. Now from the construction of $\mathcal T$, and the hypothesis $(1)$ we can say that no net in $X\setminus\{x\}$ can converge in $\{x\}$, (because if any net from $X\setminus\{x\}$ converges to {$x$} then $S$ is eventually equal to $x$, hence $S$ can't be in $X\setminus\{x\}$) i.e to say that $\{x\}$ is open. Hence $\mathcal T$ is Discrete.
Is my proof okay ? Am I missing something ? If yes then please let me know. Thank you
It's not true necessarily that $(x_i) \to x$ and the net is eventually constant than that constant must be $x$. So your proposed proof does not work as it stands.
E.g. in the space $X= \{0,1,2\}$ with open sets $\{\emptyset, \{1,2\}, X\}$ the constant net with only the value $2$ converges to $1$ and $0$ as well...
Note that this is a "local counterexample" (Lakatos' term), it shows we need a stronger assumption for the proof you propose to work:
(1) For every net $(x_i)_i$ that converges to some $x \in X$, there is some $i_0 \in I$ such that for all $i \ge i_0$: $x_i = x$ (so not just constant but constant with constant value the limit, which is stricter, but indeed holds in discrete spaces).