What shown below is a reference from "Elementos de Topología General" by Fidel Cassarubias Segura and Ángel Tamariz Mascarúa
Observation 1
Let be $J\neq\varnothing$ and let be $X:=\prod_{j\in J}X_j$ the product of not empty topological spaces. So if we pick a $x\in X$ and if we define the set
$$
\mathcal{X}_i=\{y\in X:\pi_i(y)\in X_i\wedge\pi_j(y)=\pi_j(x)\quad\forall j\neq i\}
$$
then it is possible to prove that $X_i$ is embeddable in $X$, since indeed $X_i$ is homeomorphic to $\mathcal{X}_i$.
Lemma 2
Let be $\mathcal{A}=\{A_j:j\in J\}$ a collection of connected subsets of some space $X$. So if $\bigcap\mathcal{A}\neq\varnothing$ then $\bigcup\mathcal{A}$ is connected.
Lemma 3
If $X$ is a space that contains a connected and dense subspace then it is connected.
Theorem 4
The product space $X=\prod_{j\in J}X_j$ is connected iff $X_j$ is connected for each $j\in J$.
Proof. If $X$ is connected then clearly $X_j$ is too connected, since the projection $\pi_j$ is continuous and surjective for each $j\in J$.
So now we suppose that $X_j$ is connected for each $j\in J$. So we pick an $x\in X$ and we define the set
$$
C=\{y\in X:\exists D\subseteq X\quad\text{such that is connected and such that}\quad x,y\in D\}
$$
that for the lemma 2 is connected: so if we prove that $C$ is too dense then the theorem will true for lemma 3. So we pick a basic not empty open set $A=\pi^{-1}_{j_1}(A_1)\cap…\cap\pi^{-1}_{j_n}(A_n)$ of $X$ and we prove that $A\cap C\neq\varnothing$. So for each $i=1,…,n$ we pick $a_i\in A_i$ and we define the sets
$$
C_1=\{y\in X:\pi_{j_1}(y)\in X_{j_1}\wedge\pi_j(y)=\pi_j(x)\quad\forall j\neq j_1\}
$$
and
$$
C_i=\{y\in X:\pi_{j_k}(y)=a_k\quad\text{for}\quad k=1,…,i-1\quad\text{and}\quad\pi_{j_i}(y)\in X_{j_i}\quad\text{and}\quad\pi_j(y)=\pi_j(x)\quad\text{for}\quad j\in J\setminus\{j_1,…j_{i-1}\}\}
$$
for $i\in\{2,…,n\}$. So $C_i$ is connected for each $i\in\{1,…,n\}$, since by the observation 1 it follows that $C_i$ is homeomorphic to $X_i$ for each $i\in\{1,…,n\}$. Now we observe that for each $k\in\{1,…,n-1\}$ it happens that the ponit $y\in X$ such that $\pi_{j_i}(y)=a_i\quad\text{for each}\quad i\in\{1,…,k\}$ and $\pi_j(y)=\pi_j(x)\quad\text{for each}\quad j\in J\setminus\{j_1,…,j_k\}$ is an element of $C_k\cap C_{k+1}$and so we conclude that $D:=\bigcup_{i=1}^n C_i$ is connected and $x\in D$. So it follows that $D\subseteq C$. Then it happens that $A\cap D\neq\varnothing$ and so too $A\cap C\neq\varnothing$. So we conclude that $C$ a dense subset of $X$.
For sake of completness here the original text of the proof: I hope mine was a good translation.
Well I don't understand why $C$ is connected: perhaps this is because $C$ is the union of the collection of all connected sets that contain $x$? Then I don't understand how to use the observation 1 to prove that $C_i$ is homeomorphic to $X_i$ for each $i\in\{1,…,n\}$. Then I don't understand why for each $k\in\{1,…,n-1\}$ it happens that the point $y\in X$ such that $\pi_{j_i}(y)=a_i\quad\text{for each}\quad i\in\{1,…,k\}$ and $\pi_j(y)=\pi_j(x)\quad\text{for each}\quad j\in J\setminus\{j_1,…,j_k\}$ is an element of $C_k\cap C_{k+1}$; and then why if $y\in C_k\cap C_{k+1}$ then $D$ is connected? perhaps is this for lemma 2? Finally I don't understand why $A\cap D\neq\varnothing$.
So could someone help me to understand the proof, please?
Best Answer
$C$ is indeed connected as it is a union of connected subsets that all contain $x$. So it uses your lemma 2.
$C_1$ is just $\mathcal{X}_{j_1}$ so homeomorphic to $X_{j_1}$ which is connected.
We're just constructing an intersecting "chain-path" of copies of $X_{j_1},\ldots X_{j_n}$ that will eventually contain both $x$ and a point with $a_{j_i}$ on the right coordinates, so that the union $D$ of those connected sets $C_i$ (all chain-intersecting: $C_1 \cap C_2 \neq \emptyset, C_2 \cap C_3 \neq \emptyset$, until $C_{k-1} \cap C_k \neq \emptyset$) is thus connected and shows that a point from the basic subset (with $a_{j_i}$ on the right coordinates) intersects $C$, as $D \subseteq C$ by definition of $C$.