Does there exist a connected topological space $X$ and a subset $A\subseteq X$ such that no connected component of $A$ separates $X$, but some quasi-component of $A$ separates $X$?

Meaning $X\setminus C$ is connected for every connected component $C$ of $A$, but there is a quasi-component (intersection of all relatively clopen sets of $A$ containing a particular point) $Q$ of $A$ such that $X\setminus Q$ is disconnected?

Here is my idea.

Let $S$ be the dyadic solenoid. Let $X_0$ and $X_1$ be two disjoint composants of $S$. Let $\alpha\simeq [0,1]$ be an arc in $X_0$. Let $\delta_0$ and $\delta_1$ be disjoint countable dense subsets of $\alpha$, like disjoint rationals. Define $X=(X_0\setminus \delta_0)\cup (X_1\cup \delta_1)$. The topology on $X$ will be generated by the Solenoid subspace topology together with the sets $X_0\setminus \delta_0$ and $X_1\cup \delta_1$.

$X$ is connected.

Let $A$ be a be a closed Solenoid neighborhood of $\alpha$ such that $A\neq X$. Then $\alpha\setminus \delta_0$ is contained in a quasi-component $Q$ of $A$. To see this, note that an irrational set in $\alpha\setminus (\delta_0\cup \delta_1)$ is retained by $X$, and the basic neighborhoods at these points are precisely the Solenoid neighborhoods. A sequence of arcs in $X$ uniformly limits to this set, so $$\alpha\setminus (\delta_0\cup \delta_1)\subseteq Q.$$ Since $Q$ is closed, $\alpha\setminus \delta_0\subseteq Q$.

Each connected component of $\alpha\setminus \delta_0$ is a singleton which does not separate $X$.

$X\setminus \delta_1$ is the union of two separated sets $X_0\setminus \delta_1$ and $X_1\setminus \delta_1$.

Does this sound right? The topology will only be Hausdorff, and of course ultimately I'd like regular or metrizable.