I'm getting started into this topic in topology (also in Stack Exchange) and wanted to try with a simple example.
I already know that Moore's plane is connected (I saw a proof about it with clausures), but I wanted to prove it by definition. We know it will be connected if $X = A \cup B$ (disjoint).
Therefore, by construction, the Moore plane is as a half-plane with an "extra edge" on the x-axis. In this "extra edge", each point has a "neighborhood" that extends to the right so every two open sets, no matter what will be connected by this points. However, in case this hypothesis is correct, I don't know how to write it consistently so I wonder if it's possible to explain the connection with balls with its construction.
On the other hand, to prove that it's not compact, here is my try:
Consider the collection of open sets ${U_n}$ where each $U_n$ is the set of all points (x, y) in the Moore plane such that y > n. Basically, each $U_n$ is everything above the y=n line.
This collection ${U_n}$ is an open covering of the Moore plane, because given any point (x, y) in the Moore plane, you can always find a set $U_n$ in the collection such that (x, y) is in $U_n$.
But no finite subcollection of ${U_n}$ can cover the Moore plane. For example, if you take any finite subcollection of ${U_n}$, say ${U_1, U_2, ..., U_N}$, there will always be a point in the Moore plane, like (0, N+1), that is not in any of those sets.
I will greatly appreciate any advice on being more rigorous as I have been penalized a lot before for this reason.