[proofplan] We prove connectedness by ruling out a separation. A separation would require two disjoint nonempty open subsets whose union is all of $X$. In the [indiscrete topology](/page/Indiscrete%20Topology) the only open subsets are $\varnothing$ and $X$, and these cannot form such a pair. Therefore no separation exists, so $(X,\tau)$ is connected. [/proofplan] [step:Assume a separation and use the list of open sets] Suppose, for contradiction, that $(X,\tau)$ is disconnected. By the definition of disconnectedness, there exist subsets $U,V\subset X$ such that $U$ and $V$ are open in $\tau$, both $U$ and $V$ are nonempty, $U\cap V=\varnothing$, and $U\cup V=X$. Since $\tau=\{\varnothing,X\}$, every open subset of $(X,\tau)$ is either $\varnothing$ or $X$. Because $U$ is open and $U\neq\varnothing$, we must have $U=X$. Likewise, because $V$ is open and $V\neq\varnothing$, we must have $V=X$. [guided] We argue by contradiction because the definition of connectedness is most directly negated by a separation. To say that $(X,\tau)$ is disconnected means that there are subsets $U,V\subset X$ satisfying all four separation conditions: $U$ and $V$ are open in $\tau$, the sets $U$ and $V$ are nonempty, the intersection $U\cap V$ is empty, and the union $U\cup V$ equals $X$. Now we use the defining property of the indiscrete topology. Since $\tau=\{\varnothing,X\}$, an open subset of $(X,\tau)$ has only two possibilities: it is $\varnothing$, or it is $X$. The separation condition says $U\neq\varnothing$, so the possibility $U=\varnothing$ is excluded. Hence $U=X$. The same reasoning applies to $V$: it is open and nonempty, so $V=X$. [/guided] [/step] [step:Contradict disjointness and conclude connectedness] From the previous step, $U=X$ and $V=X$. Therefore \begin{align*} U\cap V=X\cap X=X. \end{align*} The separation condition also gives $U\cap V=\varnothing$, so $X=\varnothing$. But if $X=\varnothing$, then no nonempty subsets $U,V\subset X$ can exist, contradicting the requirement that $U$ and $V$ are nonempty. Thus no separation of $(X,\tau)$ exists. Hence $(X,\tau)$ is connected. [/step]