In which of the three topologies is X compact?
Below is the way I did....
In the 1st screenshot, I showed that $X$ is compact in $T_s$, in the 2nd screenshot I showed that it is not compact in railroad metric $T_r$ and in the 3rd screenshot, I showed that it is not compact in $T_c$.
Can someone verify my proof and let me know if this looks good. If there is anything that can be changed for better style and notation, don't hesitate to let me know. Appreciate your help & support.
$(1)$ is correct, though it would be better to show explicitly that $X$ is a closed subset of $\Bbb R^2$ in the usual topology. Once you’ve done that, though, there’s no need to work as hard as you did: it immediately follows that $X$ is a compact subset and hence a compact subspace of $\Bbb R^2$ in the usual topology, and $\langle X,T_s\rangle$ is precisely that subspace.
You are correct in thinking that $\langle X,T_r\rangle$ and $\langle X,T_c\rangle$ are not compact, but there are serious problems with both of your arguments. In $(2$) the set $B\left(\langle 1,0\rangle,\sqrt2+\frac12\right)\in\mathscr{O}$ already covers $X$, so $\mathscr{O}$ does have a finite subcover. And in $(3)$ none of the sets $S_n$ is open in $T_c$: $S_n$ is open if and only if $S_n\cap S_k$ is open in $S_k$ for every $k$, but $S_n\cap S_{n+1}=\{v\}$, which is not open in $S_{n+1}$.
There is actually an easy way to show that $\langle X,T_r\rangle$ and $\langle X,T_c\rangle$ are not compact using the result in my answer to your question here. All you need is your answer to $(1)$ and the following easy result:
Theorem: Let $\langle X,\tau\rangle$ be a compact Hausdorff space, and let $\tau'$ be a topology on $X$ strictly finer than $\tau$ (i.e., $\tau\subsetneqq\tau'$). Then $\langle X,\tau'\rangle$ is not compact.
Proof: Consider the identity map $\iota:X\to X$ as a function from $\langle X,\tau'\rangle$ to $\langle X,\tau\rangle$. Since $\tau'$ is finer than $\tau$, $\iota$ is continuous. Suppose that $\langle X,\tau'\rangle$ is compact, and let $F\subseteq X$ be $\tau'$-closed; then $F$ is compact, so $F=\iota[F]$ is compact in $\langle X,\tau\rangle$ and therefore also $\tau$-closed. Thus, $\iota$ is a closed, continuous bijection and hence a homeomorphism, contradicting the hypothesis that $\tau'\ne\tau$. It follows that $\langle X,\tau'\rangle$ cannot be compact. $\dashv$