[proofplan] We prove connectedness by ruling out a separation. If two nonempty open sets $U$ and $V$ were disjoint and covered $X$, then the cofinite definition would force both complements $X\setminus U$ and $X\setminus V$ to be finite. Since $V\subset X\setminus U$, the set $V$ would be finite, and since $X\setminus V$ is also finite, this would make $X$ finite, contradicting the hypothesis. [/proofplan] [step:Assume a separation by two disjoint nonempty open sets] Suppose, for contradiction, that $(X,\tau_{\mathrm{cof}})$ is disconnected. Then there exist subsets $U,V\subset X$ such that $U$ and $V$ are open in $\tau_{\mathrm{cof}}$, both $U$ and $V$ are nonempty, $U\cap V=\varnothing$, and \begin{align*} X=U\cup V. \end{align*} [/step] [step:Use cofinite openness to force one side of the separation to be finite] Since $U$ is nonempty and open in the [cofinite topology](/page/Cofinite%20Topology), $U\neq\varnothing$, so by the definition of $\tau_{\mathrm{cof}}$ the complement $X\setminus U$ is finite. Because $U\cap V=\varnothing$, every point of $V$ lies outside $U$, hence \begin{align*} V\subset X\setminus U. \end{align*} Therefore $V$ is finite, since it is a subset of the finite set $X\setminus U$. [guided] The key point is that nonempty open sets in the cofinite topology have finite complement. We know $U$ is open in $\tau_{\mathrm{cof}}$ and $U\neq\varnothing$. By the definition \begin{align*} \tau_{\mathrm{cof}}=\{\varnothing\}\cup\{W\subset X:X\setminus W\text{ is finite}\}, \end{align*} the only [open set](/page/Open%20Set) whose complement is not required to be finite is $\varnothing$. Since $U$ is not empty, it must belong to the second part of this union. Hence $X\setminus U$ is finite. Now use disjointness. The equality $U\cap V=\varnothing$ means that no point of $V$ belongs to $U$. Equivalently, every point of $V$ belongs to the complement of $U$ in $X$, so \begin{align*} V\subset X\setminus U. \end{align*} A subset of a finite set is finite, and $X\setminus U$ is finite. Therefore $V$ is finite. [/guided] [/step] [step:Apply cofinite openness again to contradict infinitude of $X$] Since $V$ is nonempty and open in $\tau_{\mathrm{cof}}$, the definition of the cofinite topology gives that $X\setminus V$ is finite. We have already shown that $V$ is finite. Therefore \begin{align*} X=V\cup(X\setminus V) \end{align*} is a union of two finite sets, so $X$ is finite. This contradicts the hypothesis that $X$ is infinite. [/step] [step:Conclude that no separation exists] The contradiction shows that no pair of disjoint nonempty open subsets can cover $X$. Hence $(X,\tau_{\mathrm{cof}})$ is connected. [/step]