[proofplan] To prove that $(X,\tau)$ is Hausdorff, we verify the defining separation condition directly. Given two distinct points $x,y \in X$, the [discrete topology](/page/Discrete%20Topology) makes every subset of $X$ open, so the singleton sets $\{x\}$ and $\{y\}$ are open neighbourhoods of $x$ and $y$. Since $x \ne y$, these singleton neighbourhoods are disjoint, giving the required Hausdorff separation. [/proofplan] [step:Separate two distinct points by singleton neighbourhoods] Let $x,y \in X$ satisfy $x \ne y$. Since $\tau = \mathcal{P}(X)$ is the discrete topology, every subset of $X$ belongs to $\tau$. In particular, $\{x\} \in \tau$ and $\{y\} \in \tau$. The set $\{x\}$ is an open neighbourhood of $x$, and the set $\{y\}$ is an open neighbourhood of $y$. Because $x \ne y$, no element of $X$ belongs to both singleton sets, so \begin{align*} \{x\} \cap \{y\} = \varnothing. \end{align*} Thus the two distinct points $x$ and $y$ have disjoint open neighbourhoods. [guided] We prove the Hausdorff condition directly from the definitions. The Hausdorff condition asks for the following: whenever $x,y \in X$ are distinct points, there must exist open sets $U,V \in \tau$ such that $x \in U$, $y \in V$, and $U \cap V = \varnothing$. Let $x,y \in X$ satisfy $x \ne y$. Since $X$ has the discrete topology, its topology is $\tau = \mathcal{P}(X)$, the collection of all subsets of $X$. Therefore every subset of $X$ is open. In particular, the singleton subsets $\{x\}$ and $\{y\}$ are open: \begin{align*} \{x\} \in \tau \end{align*} and \begin{align*} \{y\} \in \tau. \end{align*} Now $\{x\}$ contains $x$, and $\{y\}$ contains $y$, so these are open neighbourhoods of the two points. They are also disjoint. Indeed, if some point $z \in X$ belonged to $\{x\} \cap \{y\}$, then $z = x$ and $z = y$, hence $x = y$, contradicting the assumption that $x \ne y$. Therefore \begin{align*} \{x\} \cap \{y\} = \varnothing. \end{align*} Thus for the arbitrary distinct points $x$ and $y$, we have found disjoint open neighbourhoods separating them, namely $\{x\}$ and $\{y\}$. [/guided] [/step] [step:Conclude that the discrete space is Hausdorff] The argument above applies to every pair of distinct points $x,y \in X$. Therefore, by the definition of a Hausdorff [topological space](/page/Topological%20Space), $(X,\tau)$ is Hausdorff. [/step]