Proving that if $(X,\tau)$ is infinite, then $\exists S \subset X: S \cong (\Bbb N,\tau_1)$ So, in an exercise I'm asked to prove the following:

Let $X$ be an infinite set. Then, prove that $(X,\tau)$ has a subspace homeomorphic to $(\Bbb N,\tau_1)$, where either $\tau_1$ is the trivial topology or $(\Bbb N ,\tau_1)$ is a $T_0$ - space.

So I'm having some trouble solving this because I'm not sure how to prove this kind of statements. For example: Should I assume that $\tau_1$ is trivial and then show that there exists a subspace, and then assume that $(\Bbb N,\tau_1)$ is $T_0$ and show that there also exists?
Or should I say: Let $S$ be a subspace of $(X,\tau)$ and show that it is homeomorphic to one of the two?
I don't know if I'm being able to explain myself well, but my problem is not with the concepts and the content of the proof is more with the structure of the proof. What structure should a proof for this have? I do not wish for a complete proof of this statement, I only want some help setting up the structure for the proof.
 A: We've all been there. It's confusing at first, yes.
You need to show that for every topological space, some condition holds. Then we need to take an arbitrary topological space, and show the condition holds. So let $X$ be an arbitrary topological space.
So what's the condition? If $X$ is infinite, then blah. Great, so if $X$ that we took is not infinite, there's nothing to check. So we can assume that it is infinite. Namely, we have an arbitrary infinite topological space.
Now what? Now we need to show that it has a subspace homeomorphic to $\Bbb N$ with the trivial topological, or to some $T_0$ topology on $\Bbb N$. So we have two options:

*

*If $X$ has a subspace homeomorphic to the trivial topology on $\Bbb N$, we are done. So let's assume it doesn't have one, and show that the other option holds. Or,


*If $X$ has a subspace homeomorphic to $\Bbb N$ with some $T_0$ topology, we are done. So let's assume there are none, and show that the other option holds.
Which one you choose is up to you, and it will usually depend on what you know, and how easy it is to prove each option.
A: so its look like an interesting question so i tried solving it. here is my attempt which i believe is correct.
first define an equivalence relation on X as such as $x \sim y$    if
$$ \forall\, U,V\in \tau\, where \, x\in U \,and \, y \in V\, implied \, y\in U \, and\, x \in V  $$
now its quit easy to see that it is indeed an equivalence relation, now if we have finite number of equivalence classes one of them will be an infinite set (lets call it A), and it will have the trivial topology as any open set that include one point from A will include all of them so the only open non empty set will be the entire set A so we can just choose a subset of A of size $\aleph_0$ (which can be done since A is infinite) and it will be homeomorphic to $(N,\tau_1)$ where $\tau_1$ is trivial $$$$
if however there is no such infinity class then there must be infinite numbe of class (since X itself is infinite). so now we can simply pick one point from $\aleph_0$ different classes and we will have a subset of X that is homemorphic to  $(N,\tau_1)$ that is $T_0$ since every two points in different classes there is an open set that include one and not the other from the way we defined $ \sim$. Q.E.D. now obviously the prof is not complete but i believe it can be completed quit easily
