Pseudo-connected components of the subspace $X=[0,1]\times \{\frac{1}{n}\}\cup \{(0,0),(1,0)\}$ Find the connected components $C_x$ and the $\text{pseudo-connected}^*$ components $Q_x$ of the subspace $X=[0,1]\times \{\frac{1}{n}\}\cup \{(0,0),(1,0)\}$
$* $let $X $ be a topological space and $x\in X$ the pseudo-connected component of $x$ is the set $Q_x=\cap \{A\subseteq X: x\in A, \text{where } A \:\text{is closed and open at the same time (clopen)} \}$, couldn't find the definition in English (feel free to edit it if you know it).

I will show that for every $n\in \mathbb{N}$ the set $[0,1]\times \{\frac{1}{n}\}$ is clopen.  Indeed, consider the open set $A_n=(-ε,1+ε)\times (-ε+\frac{1}{n},\frac{1}{n}+ε)$
then $A_n\cap X=[0,1]\times \{\frac{1}{n}\}$,
similarly, for the closed set $B_n=\overline{A_n}$  I get that $B_n\cap X=[0,1]\times \{\frac{1}{n}\}$
thus the set is clopen.
The set $K_n=[0,1]\times \{\frac{1}{n}\}$ is also connected because if $K_n=U\cup V$ with $U,V\in τ_X \Rightarrow U=K_i, V=K_j $ for some $i\neq j \in \mathbb{N}$
we have $\{\frac{1}i\}\cap \{\frac{1}j\}=\varnothing$
that would imply that ${[0,1]\times \{\frac{1}i\}\cup  [0,1]\times\{\frac{1}j\}=}[0,1]\times\{\frac{1}n\}$, and that cant be happening if $i,j\neq n$ so $K_n$ is connected. (I am not satisfied with that argument, but I can't explain it better).
Now because, $K_n, \: \forall n\in \mathbb{N}$ is clopen and connected we have $Q_x\subseteq K_n \subseteq C_x$
and from the lemma that says $C_x\subseteq Q_x$ we get $Q_x=C_x=K_n, \: \forall n\in \mathbb{N}  $
for the points $\{(0,0),(1,0)\}$, $C_{(0,0)=}\{(0,0)\}$ and $C_{(1,0)=}\{(1,0)\}$
Because the biggest set they can belong to is themselves, and every singleton is connected.
I can't find the  pseudo-connected components for these points, any help ?
 A: 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,
$$ \{ (0,0), (1,0)\} = Q_{(0,0)} = Q_{(1,0)}.$$
