If $G\backslash \{e\}$ is totally disconnected then $G$ is also totally disconnected? I am trying to understand the proof of the following theorem.

Theorem- If $G$ is a topological group and $G\setminus \{e\}$ is hereditarily disconnected (also known as totally disconnected), then $G$ is hereditarily disconnected.

The proof starts as : 
Obviously, every proper subgroup of $G$ is hereditarily disconnected. Therefore we may assume that $G$ is connected. 
After that, it says that since $G\backslash \{e\}$ is hereditarily disconnected, let $G\backslash \{e\}=A\cup B\ \text{where}\ A\neq 0, B\neq 0, A\cap B=\phi$ and $A$ and $B$ are separated, and hence by Chapter $5$ from Kuratowski Vol.$2$, $A\cup \{e\}$ and $B\cup \{e\}$ are connected, hence a contradiction.
Doubts- I don't understand what reasoning is applied here and why every proper subgroup of $G$ is hereditarily disconnected. I revised connectedness from Munkres but couldn't find anything that explains this reasoning. I am probably missing some important results or properties of totally disconnected spaces. I have not found much details on Totally disconnected in Munkres so if some can explain this proof with filling all the gaps, it will clear a lot of concepts for me. As I have to study more connected and totally disconnected topological groups in next few days. It will be a huge help. Thanks.
 A: No you're not missing any important result, everything is already there. Let $G$ be such that $G-\{1\}$ is totally disconnected, but not $G$. 
1) any proper subset $X$ of $G$ has a translate $g^{-1}X$ with $1\notin g^{-1}X$ (just choose $g\notin X$), so $g^{-1}X\subset G-\{1\}$ is totally disconnected, and hence so is $X$. If $G$ is not connected, this applies to the connected component of $\{1\}$. Hence $G$ is connected.
2) I don't know your "Chapter 5" says, but the reasoning is as follows: assume that $|G|\ge 3$, so $G-\{1\}$ is totally disconnected of cardinal $\ge 2$ and hence has a partition into two open subsets $A,B$. Since $B$ is open, its complement $A'=A\cup\{1\}$ is closed. Suppose by contradiction that $A'$ is not connected: then $A'=C\sqcup D$ with $C,D$ disjoint nonempty, closed in $A$, and $1\in C$. So both $C,D$ are closed in $G$. Hence $G$ is the disjoint union of nonempty closed subsets $B\cup C$ and $D$, contradiction since $G$ is connected. Hence, both $A'$ and $B'=B\cup\{1\}$ are connected.
3) Applying (1) to $A'$ and $B'$ (which are proper subsets of $G$), we deduce that $A'$, $B'$ are singletons, and reach a contradiction (when $|G|\ge 3$).
4) It remains to consider $|G|\le 2$ (for which (2) doesn't apply). Since $G$ is not totally disconnected, we have $|G|=2$, and since $G$ is connected, $G$ has the indiscrete topology. This is a counterexample (the only one!) to your theorem. 
