A space is totally disconnected if its only connected subspaces are one point sets. Show that if $X$ has the discrete topology, then $X$ is totally disconnected.
My attempt: First we show singleton sets are connected i.e. $\{ \{x\} |x\in X\}\subseteq \{A\subseteq X|$ A is connected subspace of $X\}$. Assume towards contradiction, $\{x\}$ is not connected, for some $x\in X$. So $\exists U,V\in \mathcal{T}_x=\{ T\cap \{x\}|T\in \mathcal{T}_X\}$ such that $U,V\neq \emptyset$, $U\cap V=\emptyset$ and $U\cup V=\{x\}$. Since $U,V\neq \emptyset$ and $U\cap V=\emptyset$, we have $\exists r\in U$, $\exists s\in V$ and $r\neq s$. So $r,s \in U\cup V$. Thus $|U\cup V|\geq 2$. But $|U\cup V|=1$ since $U\cup V=\{x\}$. Thus we reach contradiction. Hence $\{x\}$ is connected.
Now we show $\{ \{x\} |x\in X\}\supseteq \{A\subseteq X|$ A is connected subspace of $X\}$. $\mathcal{T}_X=\wp (X)$, power set of $X$. Assume towards contradiction, $\exists B\subseteq X$ such that $|B|\gt 1$ is connected. If $B=X$, then take $P\subset X$(proper subset) such that $P\neq \emptyset$. So $P$ and $X-P=Q$ form a separation of $X$. To make it more concrete, since $|X|\geq 2$ and $X\neq \emptyset$, $\exists x\in X$. Take $P=\{x\}$ and $Q=X-\{x\}\neq \emptyset$(because $|X|\geq 2$ ). Thus $X$ is not connected. If $B\subset X$ with $|B|\geq 2$, then take $V\subset B$ such that $V\neq \emptyset$. Since $V\subset B$, we have $V,B-V\in \mathcal{T}_B$. So $V$ and $B-V$ form separation of $B$. To make more concrete, since $|B|\geq 2$ and $B\neq \emptyset$, $\exists y\in B$. So $\{y\}$ and $B-\{y\}\neq \emptyset$ form separation of $B$. Thus $B$ is not connected. Which contradicts our initial assumption. Is this proof correct?
Edit: Inspired by Johannes Oertel. Better version of above proof: The only proper subset of singleton set is empty set. So $\nexists U\subset \{x\}$ such that $U\neq \emptyset ,\{x\}$. Only subset of $\{x\}$ is clopen is $\emptyset$ and $\{x\}$. Thus $\{x\}$ is connected vacuously.
We don’t have to consider two different cases. Assume towards contradiction, $\exists B\subseteq X$ with $|B|\geq 2$, is connected. Let $ x\in B$. Since $|B|\geq 2$, $(B-\{x\}) \neq \emptyset$. Since $\{x\},(B-\{x\})\in \mathcal{T}_X$ and $\{x\},(B-\{x\})\subset B$, we have $\{x\} ,(B-\{x\}) \in \mathcal{T}_B$. $\{x\} \cap (B-\{x\})=\emptyset$. $\{x\} \cup (B-\{x\})=B$. Thus $\{x\}$ and $(B-\{x\})$ form separation of $B$. Which contradicts our initial assumption.
Best Answer
Your attempt seems fine, although a little complicated. $X$ having the discrete topology means, that every subset is open (and hence also closed). Singletons are always connected. So if $U \subset X$ contains more than one point we have that $U = (U \setminus \{x\}) \cup \{x\}$ for any $x \in U$. Since both $\{x\}$ and $U \setminus \{x\}$ are open (and non-empty) $U$ is not connected.