Sometimes ago I have posted this question. After sometime of working I think that I have found out a different proof (not "purely topological"). I didn't post it there as an answer because (1) the following proof doesn't meet my criteria and (2) till now I am not sure if I am missing something or not.
My Proof.
Claim. Let $I$ and $J$ be two index sets. Then, $$\displaystyle\bigcup_{(\alpha,\beta)\in I\times J}(X_\alpha\times Y_\beta)=\left(\displaystyle\bigcup_{\alpha\in I}X_\alpha\right)\times \left(\displaystyle\bigcup_{\beta\in J}Y_\beta\right)$$
Proof. Observe that, \begin{align*}(x,y)\in \displaystyle\bigcup_{(\alpha,\beta)\in I\times J}(X_\alpha\times Y_\beta)&\iff(\exists (\alpha_0,\beta_0)\in I\times J)[(x,y)\in X_{\alpha_0}\times Y_{\beta_0}]\\&\iff\bigl((\exists \alpha_0\in I)[x\in X_{\alpha_0}]\bigr)\land\bigl((\exists \beta_0\in J)[y\in Y_{\beta_0}]\bigr)\\&\iff \left(x\in \displaystyle\bigcup_{\alpha\in I}X_\alpha\right)\land \left(y\in \displaystyle\bigcup_{\beta\in J}Y_\alpha\right)\\&\iff (x,y)\in \left(\displaystyle\bigcup_{\alpha\in I}X_\alpha\right)\times \left(\displaystyle\bigcup_{\beta\in J}Y_\beta\right)\end{align*}
Claim 2. Let $(X,d_X)$ and $(Y,d_Y)$ be two compact metric space. Then show that the product metric space $(X\times Y,d_{X\times Y})$ is also compact.
Proof. Let $\mathcal{F}:=\{W_\gamma:\gamma\in\Lambda\}$ be an open cover of $X\times Y$. Since $W_\gamma$ is open for all $\gamma\in \Lambda$ we conclude that for all $(x,y)\in X\times Y$ we have, \begin{align*}B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)\subseteq W_\gamma\end{align*} for some $\gamma\in \Lambda$. Consequently for all $(x,y)\in X\times Y$ we have, $$B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)\subseteq\displaystyle\bigcup_{\gamma\in\Lambda} W_\gamma$$Hence, \begin{align*}\displaystyle\bigcup_{(x,y)\in X\times Y}B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)\subseteq\displaystyle\bigcup_{\gamma\in\Lambda} W_\gamma\end{align*}Since $(X\times Y,d_{X\times Y})$ is a product metric space we have, \begin{align*}\displaystyle\bigcup_{(x,y)\in X\times Y}B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)=\displaystyle\bigcup_{(x,y)\in X\times Y}\left(B_{d_{X}}\left(x,\varepsilon_{(x,y)}\right)\times B_{d_{Y}}\left(y,\varepsilon_{(x,y)}\right)\right)\end{align*}By our previous claim we then have, $$\displaystyle\bigcup_{(x,y)\in X\times Y}\left(B_{d_{X}}\left(x,\varepsilon_{(x,y)}\right)\times B_{d_{Y}}\left(y,\varepsilon_{(x,y)}\right)\right)=\left(\displaystyle\bigcup_{x\in X}B_{d_{X}}\left(x,\varepsilon_{(x,y)}\right)\right)\times\left(\displaystyle\bigcup_{y\in Y}B_{d_{Y}}\left(y,\varepsilon_{(x,y)}\right)\right)$$Since $X$ and $Y$ both are compact, there exists finite subsets $X_0\subseteq X$ and $Y_0\subseteq Y$ such that, \begin{align*}\left(\displaystyle\bigcup_{x\in X}B_{d_{X}}\left(x,\varepsilon_{(x,y)}\right)\right)\times\left(\displaystyle\bigcup_{y\in Y}B_{d_{Y}}\left(y,\varepsilon_{(x,y)}\right)\right)=\left(\displaystyle\bigcup_{x\in X_0}B_{d_{X}}\left(x,\varepsilon_{(x,y)}\right)\right)\times\left(\displaystyle\bigcup_{y\in Y_0}B_{d_{Y}}\left(y,\varepsilon_{(x,y)}\right)\right)\end{align*}Again applying our previous claim and by properties of product metric space we get, \begin{align*}\left(\displaystyle\bigcup_{x\in X_0}B_{d_{X}}\left(x,\varepsilon_{(x,y)}\right)\right)\times\left(\displaystyle\bigcup_{y\in Y_0}B_{d_{Y}}\left(y,\varepsilon_{(x,y)}\right)\right)&=\displaystyle\bigcup_{(x,y)\in X_0\times Y_0}\displaystyle\left(B_{d_{X}}\left(x,\varepsilon_{(x,y)}\right)\times B_{d_{Y}}\left(y,\varepsilon_{(x,y)}\right)\right)\\&=\displaystyle\bigcup_{(x,y)\in X_0\times Y_0}B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)\end{align*}Since for all $(x,y)\in X\times Y$ we have $B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)\subseteq W_\gamma$ for some $\gamma\in \Lambda$, it follows that there exists a finite set $\Lambda_0\subseteq \Lambda$ such that, $$\displaystyle\bigcup_{(x,y)\in X_0\times Y_0}B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)\subseteq\displaystyle\bigcup_{\gamma\in\Lambda_0} W_\gamma$$Since, $$X\times Y\subseteq \displaystyle\bigcup_{(x,y)\in X_0\times Y_0}B_{d_{X\times Y}}\left((x,y),\varepsilon_{(x,y)}\right)\subseteq\displaystyle\bigcup_{\gamma\in\Lambda_0} W_\gamma$$we have thus obtained the finite subcover of $X\times Y$.
Question. Is my proof correct?