I have 2 questions about this proof of connected topological spaces.
Let $X$ and $Y$ be connected spaces, then $X\times Y$ is connected.
Proof. Let $p=(x_1,y_1)\in X\times Y,q=(x_2,y_2)\in X\times Y.$
Notice that $\{x_1\}\times Y\approx Y,X\times\{y_2\}\approx X.$
Hence $\{x_1\}\times Y$ and $X\times\{y_2\}$ are connected.
Also notice $\{x_1\}\times Y\cup X\times\{y_2\}$ is connected and $p,q\in\{x_1\}\times Y\cup X\times\{y_2\},$ i.e. $p$ and $q$ belong to the same component. But $p$ and $q$ were chosen arbitrarily; hence $X\times Y$ has only one component-itself.
Therefore we must have $\{x_1\}\times Y\cup X\times\{y_2\}= X\times Y,$ i.e. $X\times Y$ is connected.
What's the homeomorphism that should be consider in the proof ? and
Does the last equality is well justified?
I would write the proof in the following way:
Let $p=(x_{1},y_{1})$, $q=(x_{2},y_{2})\in X\times Y$.
$\{x_{1}\}\times Y$ is homeomorphic to $Y$, by the homeomorphism $p_{1} \colon \{x_{1}\}\times Y \longrightarrow Y$ defined as $p_{1}(x_{1},y)=y$. Similarly, $X\times \{y_{2}\}$ is homeomorphic to $X$. Then, since being connected is a topological property, $\{x_{1}\}\times Y$ and $X\times \{y_{2}\}$ are connected. Then, $\{x_{1}\}\times Y \cup X\times \{y_{2}\}$ is connected, because both are connected and the intersection is not empty ($(x_{1},y_{2})$ is in the intersection).
Finally, $X\times Y=\bigcup(\{x_{1}\}\times Y \cup X\times \{y_{2}\})$. Note that this union is also connected because of the same reason.
I hope this will help you.