Connected topology on $\mathbb Q$ such that every point is a cut point? Is there a $\text{T}_1$ topology on $\mathbb Q$ with these properties? It should be connected such removing any point disconnects the space.
 A: You should definitely doublecheck, since I’m not entirely awake, but I think that this does what you want; it’s essentially your idea, but with the topology simplified a bit. Let $X$ be the set of all finite sequence of rationals, including the empty sequence $\varnothing$. If $n\in\omega\setminus 1$, $p=\langle q_1,\ldots,q_n\rangle\in X$, $U$ is an open nbhd of $q_n$ in $\Bbb Q$, and $K$ is a compact set in $\Bbb Q$, define
$$\begin{align*}
&B(p,U,K)=\\
&\{\langle q_1,\ldots,q_{n-1},q_n',\ldots,q_m'\rangle:n\le m\in\omega\text{ and }q_n'\in U\text{ and }q_k'\notin K\text{ for }n<k\le m\}\;.
\end{align*}$$
Take 
$$\mathscr{B}(p)=\{B(p,U,K):U\text{ is an open nbhd of }p\text{ in }\Bbb Q\text{ and }K\subseteq\Bbb Q\text{ is compact}\}$$
as a local base at $p$ in $X$. For $K$ a compact subset of $\Bbb Q$ let
$$B(\varnothing,K)=\{\langle q_1,\ldots,q_n\rangle:n\in\omega\setminus 1\text{ and }q_k\notin K\text{ for }1\le k\le n\}\;,$$
and take
$$\mathscr{B}(\varnothing)=\{B(\varnothing,K):K\subseteq\Bbb Q\text{ is compact}\}$$
as a local base at $\varnothing$.
