Pseudo-connected components of the subspace $X=[0,1]\times \{\frac{1}{n}\}\cup \{(0,0),(1,0)\}$

For each $n\in \mathbb N$,

$$ X_n = X \setminus (K_1 \cup \cdots \cup K_n)$$ is clopen and contains $(0,0)$, $(1,0)$. Thus

$$Q_{(0,0)} \subset \bigcap _n X_n = \{ (0,0), (1,0)\}.$$ and similar for $Q_{(1,0)}$.

On the other hand, let $V$ is a clopen set containing $(0,0)$. Since $V$ is open, there is $N\in \mathbb N$ such that $V$ intersects $K_n$ for all $n\ge N$. Then for each $n \ge N$, $V\cap K_n$ is a nonempty clopen set in $K_n$. By connectedness of $K_n$, $V\cap K_n = K_n$ for all $n\ge N$. That is, $K_n \subset V$ for all $n\ge N$.

For any open subset $W$ containing $(1,0)$, there is $N_1$ such that $W$ intersects $K_n$ for all $n\ge N_1$. Thus $W\cap V$ is nonempty. That is, $(1,0) \in \overline V$, the closure of $V$. But since $V$ is closed, $V = \overline V$. Hence $(1,0) \in V$.

That is, every clopen subset of $X$ containing $(0,0)$ also contains $(1,0)$. Thus $\{ (0,0), (1,0)\} \subset Q_{(0,0)}$. Similar argument works for $x = (1,0)$. Hence, $$ \{ (0,0), (1,0)\} = Q_{(0,0)} = Q_{(1,0)}.$$