We consider in $\mathbb{R}^2$ the topology of the sets $U\subset \mathbb{R}^2$ which contains for each $(a,b)\in U$ a set like this $$X=\{ (x-a)^2+(y-b)^2<\varepsilon\} \cap (\{x-y =a-b\}\cup \{x+y =a+b\}),\quad \varepsilon >0$$ Study if this space verifies the first-countable axiom and the second-countably. Also, show if this space is locally compact.
I supposed that, for each $(a,b)\in \mathbb{R}^2$. I don't know how to handle this problem... Any help would be appreciate.
As mfl has observed, any set open in the Euclidean topology is open in this cross-topology. But there are many other open sets in this topology. Given a particular X, take any Euclidean open set G which contains $X \setminus \{(a,b)\}$ but such that (a,b) is a limit point of its complement. $G \cup \{(a,b)\}$ is open in the cross-topology but not in the Euclidean.
In particular, $(B((a,b),\epsilon) \setminus \{(x,y) \mid y =b \}) \cup \{(a,b)\}$ is an open nbhd of (a,b). You may notice a similarity with your previous question.
Varying the angle of the line removed will help with investigating first countability.
I'll leave local compactness for now, it's past my bedtime.
ETA Having slept on it, all becomes clear. (Not usually a tactic recommended for exams.)
Notice that if a sequence converges to (a,b) in the Euclidean topology but doesn't include any points in X, then (a,b) is not a limit point in the cross topology. The sequence is closed and not (countably) compact in that topology. You can disprove first countability and local compactness by constructing such sequences in the right places.