I'm really looking for advice on how to solve this problem, rather than the solution.
I know why any poset on a finite set has a linear extension:
It's obviously true for sets of size $0,1$ and $2$. If you have a poset on a set of size $2$, when you add a 3rd element, you add some new ordered pairs in the poset. Since you already know it holds for sets of size $2$, you can remove a minimal element and get a total order on the other $2$ elements. Then since the element that was removed is minimal, it's easy to build a total order on the entire set (let the minimal element be smaller than everything else in the total order).
So now you know that it holds for any set of size $3$, and you can easily repeat this for sets of size $4$, remove a minimal element, get a total order on the other $3$ elements and construct a total order on all $4$ elements by letting the minimal element be smaller than everything.
This strategy works for any cardinality $n$, and that's why the proof is done by induction.
The problem I have right now is that for infinitely countable sets, you can't use this strategy, you cant reach infinity by repeatedly adding $1$ element. There's no guarantee of a minimal element either. What should I do at this point? Should I try to find an answer to the question:
Why is it not possible to have a poset on a countably infinite set that cant be extended to a linear order?
I often see people say that if you are struggling with proving something, you should really understand why the theorem must actually be true.