[proofplan] We prove containment of topologies by taking an arbitrary set $U\in\tau_{\mathrm{cof}}$ and showing $U\in\tau$. The empty set is open in every topology. If $U$ has finite complement, then the $T_1$ hypothesis makes each singleton closed, and a finite union argument shows that $X\setminus U$ is closed; hence $U$ is open in $\tau$. [/proofplan]

[step:Reduce containment to an arbitrary cofinite open set] Let $U\in\tau_{\mathrm{cof}}$ be arbitrary. By the definition of $\tau_{\mathrm{cof}}$, either $U=\varnothing$ or the set \begin{align*} F:=X\setminus U \end{align*} is finite. If $U=\varnothing$, then $U\in\tau$ because every topology contains the empty set. It remains to prove $U\in\tau$ in the case where $F=X\setminus U$ is finite. [/step]

[step:Show that the finite complement is closed in the $T_1$ topology]Assume $F=X\setminus U$ is finite. If $F=\varnothing$, then $U=X$, and $U\in\tau$ because every topology contains $X$. Now suppose $F\neq\varnothing$. Choose $n\in\mathbb{N}$ and points $x_1,\dots,x_n\in X$ such that \begin{align*} F=\{x_1,\dots,x_n\}. \end{align*} Since $\tau$ is a $T_1$ topology on $X$, each singleton $\{x_i\}$ is closed in $(X,\tau)$. For each $i\in\{1,\dots,n\}$, define \begin{align*} V_i:=X\setminus\{x_i\}. \end{align*} Because $\{x_i\}$ is closed, $V_i\in\tau$. By closure of a topology under finite intersections, \begin{align*} X\setminus F=\bigcap_{i=1}^{n}V_i \end{align*} belongs to $\tau$. Since $X\setminus F=U$, we obtain $U\in\tau$.[/step]

[guided]Assume $F=X\setminus U$ is finite. Our goal is to prove that $U$ is open in the topology $\tau$. Since $U=X\setminus F$, it is enough to prove that $F$ is closed in $(X,\tau)$. First handle the degenerate finite case. If $F=\varnothing$, then \begin{align*} U=X\setminus F=X. \end{align*} Every topology contains the whole space, so $U\in\tau$. Now assume $F\neq\varnothing$. Since $F$ is finite and nonempty, there exist $n\in\mathbb{N}$ and points $x_1,\dots,x_n\in X$ such that \begin{align*} F=\{x_1,\dots,x_n\}. \end{align*} The $T_1$ hypothesis is used exactly here: in a $T_1$ topology, every singleton subset is closed. Therefore each set $\{x_i\}$ is closed in $(X,\tau)$. For each index $i\in\{1,\dots,n\}$, define the complement of the singleton by \begin{align*} V_i:=X\setminus\{x_i\}. \end{align*} Because $\{x_i\}$ is closed, its complement $V_i$ is open, so $V_i\in\tau$ for every $i$. A topology is closed under finite intersections, hence \begin{align*} \bigcap_{i=1}^{n}V_i\in\tau. \end{align*} This intersection is exactly the complement of $F$, because a point of $X$ lies in every $V_i$ precisely when it is not equal to any of the points $x_1,\dots,x_n$. Thus \begin{align*} \bigcap_{i=1}^{n}V_i=X\setminus\{x_1,\dots,x_n\}=X\setminus F=U. \end{align*} Therefore $U\in\tau$.[/guided]

[step:Conclude that the cofinite topology is contained in every $T_1$ topology] We have shown that every arbitrary $U\in\tau_{\mathrm{cof}}$ belongs to $\tau$. Therefore \begin{align*} \tau_{\mathrm{cof}}\subset \tau. \end{align*} This proves that the [cofinite topology](/page/Cofinite%20Topology) on $X$ is contained in every $T_1$ topology on $X$. [/step]