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I don't believe $[1,2)$ is open in Y, so the product topology, $X\times Y$, is then not open.

As I'm reading Topology Without Tears, I see Proposition 8.1.4 that discusses the product space being closed if the subsets of the topological spaces are closed, but nothing about when mixed.

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  • $\begingroup$ The Tychonoff product topology on $P=\prod_{j\in J}X_j$ is defined as the weakest topology on $P$ such that all the projections $p_j:P\to X_j$ are continuous. A consequence is that if $S\subset P$ and $S$ is open in the Tychonoff product topology then $p_j(S$) is open in $X_j$ for every $j\in J.$ $\endgroup$ – DanielWainfleet Dec 10 '16 at 12:52
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Because projections are open maps, $A \times B$ open in $X \times Y$ implies that both $A = \pi_X[A \times B]$ is open in $X$ and $B = \pi_Y[A \times B]$ is open in $Y$. As $B = [1,2)$ is not open in the upper limit topology (generated by sets of the form $(x,y], x < y$) (as $1$ is not an interior point of $B$), $A \times B$ cannot be open in the product.

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