Tensor products in general topology?

Let $(X,\tau)$ and $(Y,\sigma)$ be topological spaces and let $(X\times Y,\tau\times\sigma)$ be the space with the box topology. Since I never heard of it I guess that there is no space $X\otimes Y$ with a bicontinuous (that is separately continuous here, not a homeomorphism) map $\mu$ such that if $f$ is bicontinuous then there exist an unique continuous map $\varphi$ making the diagram commute. $\require{AMScd}$ \begin{CD} X\times Y @>\mu>> X\otimes Y\\ @V f V V\# @VV \exists!\varphi V\\ Z @= Z \end{CD} How to prove that such a $X\otimes Y$ in general not exists?

The general notion of a "bimorphism" and the classifying tensor product has been studied in the paper

B. Banaschewski and E. Nelson. Tensor products and bimorphisms. Canad. Math. Bull, 19(4):385-402, 1976.

It contains a general theorem which shows that tensor products "almost always" exist. In the case of topological spaces, the construction looks as follows:

First, consider the coproduct $P=\coprod_{x \in X} Y \sqcup \coprod_{y \in Y} X$. For each $x \in X$ we have an inclusion $i_x : Y \to P$. For each $y \in Y$ we have an inclusion $j_y : X \to P$. Define $X \otimes Y$ to be the quotient space $P/{\sim}$ where $i_x(y) \sim j_y(x)$. This obviously satisfies the required universal property. Notice that the underlying set of $X \otimes Y$ is just $\{(x,y) : x \in X, y \in Y\}$, and the topology looks as follows: A subset $U$ is open iff for every $x \in X$ the set $\{y \in Y : (x,y) \in U\}$ is open in $Y$ and for every $y \in Y$ the set $\{x \in X : (x,y) \in U\}$ is open in $X$.

If we apply this type of construction to the category of modules, we also obtain an alternative construction of the tensor product of modules (SE/291644).