Here is my attempt:
''Let $\mathcal{G}$ be a open cover for $X\times Y$. Let $y$ in $Y$, since $X \times {y} \cong X$, then $X \times {y}$ is compact. Then, there is finite subcover $\mathcal{G}_{y}=\{G_{y,1},G_{y,2}, \dots, G_{y,n}\}$ of $\mathcal{G}$ such that $$X \times {y}=\bigcup \mathcal{G}_{y} = \bigcup_{i=1}^{n}G_{y,i}. $$ Now $\{\pi_{2}(\mathcal{\bigcup G}_{y}) \ |\ y \in Y\}$ (where $\pi_{2}:X\times Y \to Y$ is projection) is a open cover of $Y$. Since $Y$ is Lindelöf, there is countable subcover of $\{\pi_{2}(\mathcal{\bigcup G}_{y}) \ |\ y \in Y\}$, say $\{\pi_{2}(\mathcal{\bigcup G}_{y_{i}}) \ |\ i \in \mathbb{N}\}$. Thus, $\{\bigcup \mathcal{G}_{y_{i}} \ | \ i\in \mathbb{N}\}$ is countable subcover of $\mathcal{G}$. Hence, $X\times Y$ is Lindelöf.''
Is it true? If it is true, why can't similar proof be done for Tychonoff's theorem?
Your argument is faulty. From$$ \pi_2\left(\bigcup_{i\in\mathbb N}\mathcal G_{y_i}\right)= Y $$ we cannot conclude $$ \bigcup_{i \in \mathbb N}\mathcal G_{y_i} = X \times Y . $$
A simple example. Let $X \times Y$ be the square $[0,1] \times [0,1]$ shown in gray.
Let $A$ be a triangle shown in yellow. Let $B$ be a square shown in blue. Let $C$ be a triangle shown in green. Let $D$ be a square shown in red.
Suppose we have chosen (among others): $\mathcal G_0 = \{A,B\}$ which covers $X \times \{0\}$; and $\mathcal G_1 = \{C,D\}$ which covers $X \times \{1\}$. [Of course we have also chosen $\mathcal G_y$ for $0<y<1$, which we will not need below.]
Then we get (approximately) $\pi_2(\bigcup \mathcal G_0) = [0,9/10)$ and $\pi_2(\bigcup \mathcal G_1) = (1/10,1]$. So these two sets cover $Y$. [That's why we don't need the other $\mathcal G_y$.]
But $\bigcup \mathcal G_0 \cup \bigcup \mathcal G_1$, shown below, is not $X \times Y$. There is still some gray showing.