I was reading James Munkre's Example on the counterexample that a product of two quotient maps is not a quotient map,
Let $X = \mathbb{R}$ and $X^{*}$ be the quotient space obtained from $X$ by identifying the subset $\mathbb{Z}_{+}$ to a point $b$; let $p: X \rightarrow X^{*}$ be the quotient map. Let $\mathbb{Q}$ be the subspace of $\mathbb{R}$ constitng of the rational numbers; let $i: \mathbb{Q}\rightarrow \mathbb{Q}$ be the identity map. We show that $$ p \times i : X \times \mathbb{Q} \rightarrow X^{*} \times \mathbb{Q}$$ is not a quotient map.
I was wondering over here that why the following argument is not valid (which tries to show that it is in fact a quotient map):
Note that $p \times i $ is a continuous surjection, since if $U \times V$ is open, then $$ ( p \times i )^{-1} ( U \times V) = p^{-1}(U) \times i^{-1}(V) = p^{-1}(U) \times V$$ which is open in $X \times \mathbb{Q}$.
Also we have $(p \times i )^{-1} ( A \times B) = p^{-1}(A) \times i^{-1}( B) $, so if its an open set, then both $p^{-1}(A)$ and $i^{-1}(B)$ are arbitrary union of open sets of $X$ and $\mathbb{Q}$. Further, since both $p$ and $i$ are quotient maps, $A$ and $B$ are open in $X^{*}$ and $\mathbb{Q}$ respectively. Hence, $A \times B$ is open in $X^{*} \times \mathbb{Q}$.
Thus I have showed that $(p \times i)$ is a continuous surjection and it maps saturated open sets to open set, hence a quotient map(??)
As PeterJL noted in the comments, you’ve not shown that $p\times i$ maps saturated open sets to open sets (or, equivalently, maps saturated closed sets to closed sets). And in fact it does not, as the following example shows.
For $k,n\in\Bbb Z^+$ with $k\ge 2$ let
$$I_{n,k}=\left[\frac{\sqrt2}n-\frac1k,\frac{\sqrt2}n+\frac1k\right]\cap\Bbb Q\;,$$
and let
$$F_{n,k}=\left\{n-\frac1k\right\}\times I_{n,k}\;;$$
clearly $F_{n,k}$ is closed in $X\times\Bbb Q$. Let
$$F_n=\bigcup_{k\ge 2}F_{n,k}\;.$$
Clearly
$$\bigcap_{k\ge 2}\left[\frac{\sqrt2}n-\frac1k,\frac{\sqrt2}n+\frac1k\right]=\left\{\frac{\sqrt2}n\right\}\;,$$
and $\frac{\sqrt2}n$ is irrational, so $\bigcap_{k\ge 2}I_{n,k}=\varnothing$, and hence $F_n$ is closed. Finally, let $F=\bigcup_{n\in\Bbb Z^+}F_n$; it’s not hard to check that $F$ is also closed.
$F$ is disjoint from $\Bbb Z^+\times\Bbb Q$, so $F$ is automatically saturated, and $(p\times i)[F]=F$. Let
$$p=(p\times i)(\langle 1,0\rangle)\;,$$
the point of $X^*\times\Bbb Q$ whose preimage under $p\times i$ is $\Bbb Z^+\times\{0\}$. Suppose that $U$ is an open nbhd of $p$ in $X^*\times\Bbb Q$; then $(p\times i)^{-1}[U]$ is an open nbhd of $\Bbb Z^+\times\{0\}$ in $X\times\Bbb Q$. Let $V=(p\times i)^{-1}[U]$; $V$ is not just open, but also saturated, so there is an open nbhd $W$ of $0$ in $\Bbb Q$ such that
$$V\cap\big(\{n\}\times\Bbb Q\big)=\{n\}\times W$$
for each $n\in\Bbb Z^+$. We may assume that $W=\left(-\frac1m,\frac1m\right)\cap\Bbb Q$ for some $m\in\Bbb Z^+$.
Now choose $n>m$ large enough so that $\frac{\sqrt2}n<\frac1m$; $\langle n,0\rangle\in V$, so there is an $\epsilon>0$ such that
$$(n-\epsilon,n+\epsilon)\times W\subseteq V\;.$$
Choose $k\ge 2$ such that $\frac1k<\epsilon$; then
$$F\cap V\supseteq F_{n,k}\cap V\supseteq F_{n,k}\cap\big((n-\epsilon,n+\epsilon)\times W\big)\ne\varnothing\;.$$
That is, every saturated open nbhd of $\Bbb Z^+\times\{0\}$ in $X\times\Bbb Q$ has non-empty intersection with $F$, and it follows that $p\in\big(\operatorname{cl}_{X^*\times\Bbb Q}F\big)\setminus F$ and hence that $F=(p\times i)[F]$ is not closed in $X^*\times\Bbb Q$, even though $F$ is saturated in $X\times\Bbb Q$.