Successor topology EDIT: I answered it myself below, please upvote it if you agree.
Let $X = \mathbb{N}$ and let the topology equipped be given via
$$\tau=\{U \subset \mathbb{N} : (2n-1) \in U \Rightarrow 2n \in U\}$$
Show $(X,\tau)$ is locally compact but not compact.
For locally compact, I need to have a compact set containing an open set containing any given $x \in \mathbb{N}$
So does this split into cases where $x \in \mathbb{N}$ is even or is odd? If $x \in \mathbb{N}$ is even, can we take my open neighborhood to be $\{x\}$ and if $x \in \mathbb{N}$ is odd then $\{x,x+1\}$ can be my $U$? and my compact set containing it can be $\{x,x+1,x+2,x+3\}$?
And for not compact do I take an arbitrary open covering, say
$$\mathbb{N} \subseteq \bigcup_{\alpha \in A} U_\alpha$$
where $A$ is an arbitrary indexing set, and show it has no finite sub covering? could I do this by contradiction? Suppose there exists a finite subset $B \subset A$ such that
$$\mathbb{N} \subset \bigcup_{\beta \in B} U_\beta$$
Am I on the right path?
 A: For locally compact:
Let $x \in \mathbb{N}$, if $x$ is even, then take
$$U:=\{x\}$$
and to find a set properly containing it take our compact set to be the finite set
$$K:=\{x,x+1\}.$$
If $x$ is odd, then take
$$U:=\{x,x+1\}$$
and
$$K:=\{x,x+1,x+2\}$$
Then we have found a compact set containing the open set containing each point of $\mathbb{N}$, and thus $(X,\tau)$ is locally compact.
For not compact, take the open cover
$$\bigcup_{n \in \mathbb{N}} \{2n-1,2n\}$$
which has no finite sub cover. If it did, then there would exists a finite subset $A \subset \mathbb{N}$ such that
$$\mathbb{N} \subset \mathcal{A}= \bigcup_{n \in A}\{2n-1,2n\}$$
But then as $\mathbb{N}$ is infinite, there exists an $N \in \mathbb{N}$ such that
$$N=\max_{n \in A}+1$$
and
$$\{2N-1,2N\} \not\in \mathcal{A}$$
thus no finite sub cover exists and $(X,\tau)$ is not compact, as needed.
A: It is locally compact because ${2n-1, 2n}\$ is an open cover made of finite sets.
It is not compact because, for example:

*

*The open cover $\{2n-1, 2n\}$ is an infinite open cover without subcover; or

*The set of odd numbers is an infinite closed set (since the even are open) with discrete topology (therefore, not compact).

