The equivalence relation induced by the partition of the quasi-component is multiplicative. Definition
The quasi-component $Q_x$ of $x\in X$ is the intersection of the clopen set that contains $x$.
Observation
Since the intersection (infinite as well) of closed sets is always closed,  any quasi-component is a closed set.
Lemma
The quasi-components of two distinct points of a topological space $X$ either coincide or are disjoint, so that all quasi-components constitute a decomposition of the space $X$ into pairwise disjoint closed subsets.
Proof. So if $y\in Q_x$ for some $x,y\in Q_x$ then clearly $Q_y\subseteq Q_x$ and so this means that if $Q$ is clopen and $x\in Q$ then $y\in Q$. So now we suppose that $Q_y\subset Q_x$, that is there exist a clopen set $Q$ such that $Q\subseteq X\setminus\{x\}$ and so $X\setminus Q$ is a clopen such that $x\in X\setminus Q$ and so $y\in X\setminus Q$, but this would mean that $Q\cap X\setminus Q\neq\varnothing$ and clearly this is impossible.
Corollary
For any topological space $X$ the relation 
$$
x\approx y\iff Q_x=Q_y
$$
for any $x,y\in X$ is an equivalence relation in $X$.
Corollary
For any topological space $X$ and for any $x,y\in X$ it follows that $x\approx y$ iff for any binary open partition of $X$ it follows that $x$ and $y$ belong to the same part.
Proof. If $x\approx y$ clearly then if there exist a binary partition of $X$ such that $x$ and $y$ don't belong to the same part then there exist two disjoint open set $U$ and $V$ such that $x\in U$ and $y\in V$ and $U\cap V=\varnothing$; but if this was happen then $U$ and $V$ are two clopen set such that $x\in U$ and $y\in V$ and $x\notin V$ and $y\notin U$ and this would be inconsistence respect to the assumption $x\approx y$. Now we observe that if $Q\subseteq X$ is clopen then $Q$ and $X\setminus Q$ induce  a open partition on $X$. So if $x,y\in X$ are such that for any binary partition of $X$ it follows that $x$ and $y$ belong to the same part then if $Q$ is a clopen set and $x\in Q$ then $y\in Q$ and so clearly $Q_x=Q_y$.
Now what shown below is a theorem of "Topology II" by Kazimierz Kuratowski

So I don't understand the point $2$ of the proof and so to explain it I proved the second corollary: so is its proof correct? then using it could I explain the point 2? Then I don't understand why $\mathfrak{z\approx m}\in(\mathfrak{z^1,...,z^n})\times\mathscr{X}_n$ and $\mathfrak{m\approx y}\in\mathscr{X_1\times...\times X_n}\times\mathfrak{y}^n$. Then why $\mathfrak{z\approx m}\in\Big((\mathfrak{z^1,...,z}^{(n-1)})\times\mathscr{X_n}\Big)\cup\Big(\mathscr{X_1\times...\times X}_{n-1}\times(\mathfrak{y}^n)\Big)$ and so why $\mathfrak{z\approx m}$ in $\mathscr{X_1\times...\times X}_n$? So if this is true why $\mathfrak{z\approx y}$ in $\mathscr{X_1\times...\times X}_n$? Finally why if $(\mathfrak{z}^{t_1},...\mathfrak{z}^{t_n})\approx(\mathfrak{y}^{t_1},...,\mathfrak{y}^{t_n})$ in $\mathscr{X_{t_1}\times....\times X_{t_n}}$ then $\mathfrak{m\approx y}$ in $\mathscr{R}$ and so then why in $\mathscr{X}$?
So could someone help me, please?
 A: The connection between $Q_x$ and $\approx$, i.e.
$$Q_x = Q_y \iff x \approx y\tag{a}$$
I would show as follows: suppose $x \approx y$. Then let $z \in Q_x$; we want $z \in Q_y$, so let $C$ be any clopen of $X$ that contains $y$. As $C,C^\complement$ partitions $X$, and $x \approx y$, we get $x \in C$. So $z \in Q_x \subseteq C$, so $z \in C$; as $C$ was arbitrary, $z \in Q_y$. So $Q_x \subseteq Q_y$, and by a symmetrical argument (interchange $x$ and $y$) we get $Q_y \subseteq Q_x$ and the left hand side of $(a)$ has been shown. Now suppose $Q_x=Q_y$, and let $U,V$ be a (cl)open partition of $X$ with $x \in U$ (say). As $y \in Q_y = Q_x \subseteq U$, $y \in U$ as well, and as the partition was arbitrary, $x \approx y$. 

As to productivity of $\approx$, we want to show in $X=\prod_{j \in J} X_j$ that
$$x=(x_j)_j \approx y=(y_j)_j \iff \forall j \in J: x_j \approx y_j\tag{1}$$
The point $(2)$ is to see the left to right implication of $(1)$. If two points are equivalent but in some coordinate $x_j \not\approx y_j$ ,we can separate them by a (cl)open partition $G_j, H_j$ of $X_j$ and then $\pi_j^{-1}[G_j]$ and $\pi_j^{-1}[H_j]$ form a (cl)open separation between them in the product, contradiction. That's all he's saying there;
It's also the general argument that $$Q_x \subseteq \prod_j Q_{x_j}$$
as it were.
The finite case hinges on the case $n=2$ really: if $x_1 \approx x_2$ in $X$ and $y_1 \approx y_2$ in $Y$, the claim is that $(x_1,x_2) \approx (y_1,y_2)$ in $X \times Y$ (and then induction implies the finite product equality case).
Kuratowski uses that $\{x_1\} \times Y \simeq Y$ (homeomorphism via projection) and $X \times \{y_2\} \simeq X$ as well, and $(x_1,y_2)$ is their "link": if $C_1, C_2$ is a binary clopen partition of $X \times Y$, their intersection with  $X \times \{y_2\}$ is one too, and this contains both $(x_1,y_2)$ and $(x_2,y_2)$ and as $x_1 \approx x_2$ and these correspond under the homeomorphism, we see that $(x_1, y_2)$ and $(x_2,y_2)$ are in the same clopen set, say $C_1$ for definiteness. 
Similarly $y_1 \approx y_2$ in $Y$ so in the homeomorphic space $\{x_1\} \times Y$ we know that $(x_1, y_1)$ and $(x_1, y_2)$ are in the same clopen set (in the partition of $C_1,C_2$ intersected with $\{x_1\} \times Y$) and we already know the $(x_1,y_2)$ is in $C_1$ so $(x_1,y_1)$ is too. In conclusion, $(x_1,y_1)$ and $(x_2,y_2)$ are both in $C_1$, and as the partition was arbitrary, $(x_1,y_1) \approx (x_2,y_2)$.
So in the finite case (by obvious induction, using $(X_1 \times \ldots X_{n_1}) \times X_n \simeq X_1 \times \ldots X_{n-1} \times X_n$, etc.)
$$(x_1,\ldots, x_n) \approx (y_1, \ldots,y_n) \iff \forall 1 \le i \le n: x_1 \approx y_i$$
Note the similarity with the usual proof of connectedness of finitely many connected spaces, using a glueing argument. 
The general case uses the "finiteness" of product open basis sets in a similar way: we assume that $\forall j \in J: x_j \approx y_j$ and want to show $x=(x_j)_j \approx (y_j)_j=y$ (in modern notation). So let $G,H$ be a binary clopen partition of $X =\prod_j X_j$ and say $x \in G$. We can find a finite set of indices $F \subseteq J$ such that $O= \prod_j O_j$ obeys that $O_j = X_j$ for $j \notin F$ and all $O_j$ are open in $X_j$ and $$x \in O \subseteq G$$
Then Kuratowski forms the auxiliary point $(w=(w_j)_j$ defined by 
$$w_j=\begin{cases} x_j & j \in F\\ y_j &j \notin F\end{cases}$$
and note that $w \in O$ so $w \in G$. 
Also $X_F:=\prod_{j \in J} Y_j \subseteq X$, with $Y_j= X_j, j \in F$, $Y_j=\{y_j\}, j \notin F$, is homeomorphic to $\prod_{j \in F} X_j$, a finite product and contains $w$ and $y$, with $\pi_F$ as a homeomorphism, and $x_j \approx y_j$ for $j \in F$ in particular, the finite case we just showed then implies that $(x_j)_{j \in F} \approx (y_j)_{j \in F}$ too, so their homeomorphic images $w \approx y$ in $X_F$, so $y \in G$ too (using the induced partition of $G \cap X_F,H \cap X_F$ again). Hence $x \approx y$ and we're done.
