A finer topology making a connected set disconnected How to find an example of a topological space $(X,\tau)$ such that it is connected but when we consider a finer topology $\tau^{'}$on the same set $X$,then we get $(X,\tau ^{'})$ to be disconnected?
 A: As Martín-Blas Pérez Pinilla points out in a comment, there is a very easy example.  Suppose $\langle X, \tau\rangle$ is connected.  Then $\tau$ is a subset of the power set of $X$  by definition, and the power set of $X$ is a topology for $X$, called the discrete topology. It is easy to show that if $X$ is given the discrete topology it is disconnected unless it contains fewer than two points.
I would like to add that there is an important pattern of reasoning here. The question asks you to show that $X$ can become disconnected if we make its topology finer.  It should be clear that while making the topology finer can turn a connected space into a disconnected space, the opposite never happens: you cannot turn a disconnected space into a connected space by making the topology finer. This is because $X$ is disconnected if we can find a partition of $X$ into open sets, and if these sets are open in one topology, they are open in a finer topology because this is the definition of “finer”.
So the thing to try is the finest possible topology, and if that topology does not make $X$ disconnected, then nothing can. The finest possible topology is the discrete topology, and that does solve the problem.
A: If $(X,\tau)$ is disconnected, it means that $X=A\cup B$, where $A$ and $B$ are disjoint, each one open and closed. If $\tau'$ is a finer topology, then both $A$ and $B$ are still open and closed in $\tau'$, which makes $(X,\tau')$ disconnected. Thus, the above described situation is impossible.
A: Suppose we have a connected space $(X,\mathcal{T})$, where $X$ has more than 1 point. Then the topology $\mathcal{T}$ is not discrete (or else we'd have disconnected space), so there is some $D \subset X$ such that $D \notin \mathcal{T}$. Then let $\mathcal{T'}$ be the smallest topology that contains $\mathcal{T} \cup \{D, X \setminus D\}$ (i.e. the intersection of all topologies that has this as a subset). Then $\mathcal{T}'$ is finer than $\mathcal{T}$ and $X$ is disconnected in that topology, as $\{D, X \setminus D\}$ is a disconnection of $X$. 
So any connected topology has lots of disconnected finer topologies. 
A: Late but can be useful for future readers!
Consider $(X, \tau_{\text{indiscrete}})$ , the only non empty open set is $X$ itself. Hence $(X, \tau_{\text{indiscrete}})$ is connected.
Indiscrete topology is the smallest (coarset ) topology that one can defined on a set i.e any topology $\tau$ on $X$ is finer than $\tau_{\text{indiscrete}}$.
Now choose a  topology $\tau$ on which is strictly finer than indiscrete topology such that $(X, \tau) $ is disconnected.
For $|X|\ge 2$ (otherwise every topology on $X$ is nothing but indiscrete topology), one can choose discrete topology on $X$ which is clearly disconnected ( in fact totally disconnected).

There are lot's of examples. I want to mention one particular example :
Consider $\Bbb{R}$ and two topologies $\tau_{\text{std}}$ and $\tau_{\text{Lower limit}}$ which are generated by the basis
$\mathcal{B}_{\text{std}}=\{(a,b):a,b\in\Bbb{R} ,a<b\}$
$\mathcal{B}_{\text{Lower limit}}=\{[a,b):a,b\in\Bbb{R} ,a<b\}$
respectively.
Clearly $\tau_{\text{std}}\subsetneq\tau_{\text{Lower limit}}$
as $(a, b) =\bigcup_{n\in\Bbb{N}}[a+\frac{1}{n},b)\in \mathcal{B}_{\text{Lower limit}}$

$(\Bbb{R}, \tau_{\text{std}}) $ is connected. But $(\Bbb{R}, \tau_{\text{Lower limit}}) $ is not connected ( in fact totally disconnected). As $$\Bbb{R}\setminus [a, b) =(-\infty, a) \cup [b, \infty) \in (\Bbb{R}, \tau_{\text{Lower limit}}) $$ Hence $[a, b) $ is a clopen set.
