Let $\mathbb Q$ be the set of rationals with induced euclidian topology, let $\sim_1$ be the relation on $\mathbb Q$ which identifies all the integers, and let $\sim_2$ be the identity relation on $\mathbb Q$.
Why the spaces $(\mathbb Q\times\mathbb Q)/({\sim}_1\times{\sim}_2)$ and $(\mathbb Q/{\sim}_1)\times(\mathbb Q/{\sim}_2)$ are not homeomorphic?
The space $X=(\mathbb{Q}\times\mathbb{Q})/({\sim}_1\times{\sim}_2)$ is a quotient of a sequential space and hence sequential, so it suffices to show the space $Y=(\mathbb{Q}/{\sim}_1)\times(\mathbb{Q}/{\sim}_2)$ is not sequential. Fix a sequence $(\alpha_n)$ of positive irrational numbers converging to $0$. For each $n\in\mathbb{N}$, choose a sequence of points $(x_{mn},y_{mn})\in \mathbb{Q}^2$ converging to $(n,\alpha_n)$ such that $0<|x_{mn}-n|<1/2$ for all $m$. Let $A=\{(x_{mn},y_{mn}):m,n\in\mathbb{N}\}$, thought of as a subset of $Y$ in the obvious way. Then I claim that $A$ is sequentially closed but not closed in $Y$, so $Y$ is not sequential.
First, $A$ is not closed: it is easy to see that any open neighborhood of the point $(\mathbb{Z},0)$ in $Y$ must intersect $A$ because $\alpha_n$ converges to $0$ (here $\mathbb{Z}$ denotes the equivalence class of the integers as an element of $\mathbb{Q}/{\sim}_1$). Since $(\mathbb{Z},0)\not\in A$, $A$ is not closed.
Second, $A$ is sequentially closed. In fact, I claim that no sequence of distinct elements of $A$ converges in $Y$. Let $(a_k)$ be a sequence of distinct elements of $A$; then there are integers $m_k$ and $n_k$ such that $a_k=(x_{m_kn_k},y_{m_kn_k})$ for each $k$. Suppose that $(a_k)$ converges; then $(x_{m_kn_k})$ must converge in $\mathbb{Q}/{\sim}_1$ and $(y_{m_kn_k})$ must converge in $\mathbb{Q}$. Since the numbers $\alpha_n$ are irrational, the only way $(y_{m_kn_k})$ can converge is if the sequence $(n_k)$ is unbounded. Passing to a subsequence, we may assume that the integers $n_k$ are all distinct. But now I claim the sequence $(x_{m_kn_k})$ cannot converge. Indeed, it is clear that the only point it could possibly converge to is $\mathbb{Z}$. For each $n\in\mathbb{Z}$, let $U_n\subset\mathbb{Q}$ be an interval around $n$ small enough to not contain $x_{m_kn_k}$ for any $k$ such that $n_k=n$ (there is at most one such $k$). Then the image of $U=\bigcup U_n$ in $\mathbb{Q}/{\sim}_1$ is a neighborhood of $\mathbb{Z}$ that does not contain any $x_{m_kn_k}$. Thus $(x_{m_kn_k})$ cannot converge to $\mathbb{Z}$, and hence cannot converge at all. This is a contradiction, so $(a_k)$ cannot converge.