[proofplan] We prove the equivalent formulation of total disconnectedness: every connected subset has at most one point. Let $A \subset X$ contain two distinct points. Using the [discrete topology](/page/Discrete%20Topology), we separate $A$ into the singleton of one point and its complement inside $A$. This gives a separation of $A$, so no subset with two distinct points is connected. [/proofplan] [step:Separate any subset containing two distinct points] Let $A \subset X$ be a subset containing two distinct points. Choose $x,y \in A$ with $x \neq y$. Define two subsets of $A$ by \begin{align*} U := \{x\}, \qquad V := A \setminus \{x\}. \end{align*} Since $\tau = \mathcal{P}(X)$, both $\{x\}$ and $X \setminus \{x\}$ are open in $(X,\tau)$. Therefore $U = A \cap \{x\}$ and $V = A \cap (X \setminus \{x\})$ are open in the [subspace topology](/page/Subspace%20Topology) on $A$. The sets $U$ and $V$ are disjoint, their union is $A$, and both are nonempty: $x \in U$ and $y \in V$. Hence $A$ is disconnected. [guided] We want to show that a subset with at least two points cannot be connected. Let $A \subset X$ contain two distinct points, and choose $x,y \in A$ with $x \neq y$. The natural way to disconnect $A$ is to isolate one point from the rest of the set. Define \begin{align*} U := \{x\}, \qquad V := A \setminus \{x\}. \end{align*} Because $X$ has the discrete topology, every subset of $X$ is open. In particular, $\{x\}$ is open in $X$, and $X \setminus \{x\}$ is also open in $X$. The subspace topology on $A$ declares a subset of $A$ to be open exactly when it is the intersection of $A$ with an open subset of $X$. We have \begin{align*} U = A \cap \{x\} \end{align*} and \begin{align*} V = A \cap (X \setminus \{x\}). \end{align*} Thus both $U$ and $V$ are open in the subspace topology on $A$. Now we check the separation conditions. The sets are disjoint because no point can both equal $x$ and fail to equal $x$. Their union is $A$ because every point of $A$ either equals $x$ or does not equal $x$. Finally, both pieces are nonempty: $x \in U$, while $y \in V$ since $y \in A$ and $y \neq x$. Therefore $A$ is the union of two disjoint nonempty subsets that are open in the subspace topology on $A$, so $A$ is disconnected. [/guided] [/step] [step:Conclude that every connected subset has at most one point] The preceding step proves that every subset $A \subset X$ with at least two points is disconnected. Therefore, if $C \subset X$ is connected, then $C$ cannot contain two distinct points. Hence every connected subset of $(X,\tau)$ has at most one point. By the definition of a totally disconnected space in this equivalent form, $(X,\tau)$ is totally disconnected. [/step]