This is related to an earlier problem I submitted on this site. Let $Y \subset \mathbb{R}^2$ be the x-axis.
I'm trying to figure out what the subspace topology on $Y$ will be if $\mathbb{R}^2$ has the product topology $\mathbb{R}_u \times \mathbb{R}_u$, where $\mathbb{R}_u$ is the upper limit topology.
In this case, the product topology in question will have a basis of the form $\mathcal{B} = \{(a,b] \times (c,d] | a,b,c,d\in\mathbb{R}, a<b,c<d\}$.
We know if $\mathcal{B}$ is the basis for a topology on $\mathbb{R}^2$, then $\mathcal{B}_Y = \{B \cap Y | B \in \mathcal{B}\}$ is the basis for the subspace topology on Y.
Hence the basis of this topology on $\mathbb{R}^2$ will be all sets of elements of the form $Y \cap ((a,b]\times (c,d])$.
$Y$ is the x-axis, so $(y,0)$ for all elements of the subset $Y$.
I have a little confusion here. I believe $(a,b]\times (c,d]$ is a closed subset of $\mathbb{R}^2$ because the product of closed sets is closed and $(a,b]$ is closed on $\Bbb R$. I think that any intersection with $Y$ is still $(a,b]\times\{0\}$ because the value of $Y$ has not changed.
If this is true, then you could proceed as below. If it's not, then I'm baffled about how you treat closed sets in this case.
Based on ideas in the previous problem, having $(a,b]$ as an open subset of $Y$ tells us that for an open subset $A$ of $Y$, $A=Y\cap A^\star$ for some open subset $A^*$ of $\mathbb{R}^2$.
If $(y,0)\in A$, then $(y,0)\in A^\star$. We can then assume $(y,0)\in(a,b]\times(c,d]$ for some value of $a,b,c,d$ and that $(a,b]\times(c,d]\subset A^\star$. We then say $c<0<d$ and $\bigl((a,b]\times(c,d]\bigr)\cap Y=(a,b]$.
This give us $A=\bigcup_{y\in Y}(a,b]$, so we know that $A$ belongs to the upper limit topology.
Hence the inherited topology is the upper limit topology.
I think this same concept would also apply to a $Y \subset \mathbb{R}^2$ where $\Bbb R^2$ has the product topology $\Bbb R_l \times \Bbb R_l$, I chose $\Bbb R_u$ simply for the connection to the last question asked.
In any product $X \times Y$ a set like $X \times \{y_0\}$ is homeomorphic to $X$, simply by the projection $\pi: X \times \{y_0\} \to X$ and its continuous inverse $x \to (x,y_0)$ which is just the identity times a constant map... The same holds with $Y$ and sets like $\{x_0\} \times Y$, with $x_0 \in X$.
The "axes" are just homeomorphic to the factor space the run parallel to.
So indeed the $x$-axis essentally has the upper-limit topology, and the $y$-axis as well. If $X=Y$ the diagonal $\Delta_X$ (the line $y=x$) also is homeomorphic to $X$...