A few questions on $\omega_1$ I'm trying to understand several things regarding $\omega_1$ and trying to get a better feeling of it.
First question is - is $\omega_1$ a connected space? I think it isn't, but not really sure how to prove it. Can I just say that it's the union of all the smaller ordinals? 
Another thing I'm not sure about is whether it is path-connected? 
3$^{rd}$ question is, taking the linear order on $\omega_1$, does it have a countable chain condition?
If it's not connected, can it still be compact? Or does it go the other way - if it's not compact it can't be connected? 
And I think I read somewhere it is not separable, but I'm not sure why?
I'm sorry if my questions sounds a bit "dumb", I'm just new to this, and trying to get a better "feeling" of how things go.
 A: $\omega_1$ has isolated points (any successor ordinal, for example), therefore it's not connected, and therefore not path-connected.
Connectedness doesn't have a lot to do with compactness. There are connected spaces that aren't compact, there are compact spaces that aren't connected. There are even compact spaces with an infinite number of connected components (the Cantor space for example; such a space is necessarily not locally connected, though).
The best way to prove that it's not compact is to exhibit an open cover that doesn't have a finite subcover. By definition of the topology of $\omega_1$, the sets $X_\alpha = \{ \beta \in \omega_1 : \beta < \alpha \}$ (where $\alpha < \omega_1$) are open, and then $\{ X_\alpha \}_{\alpha \in \omega_1}$ is an open cover (because every $\beta \in X_{\beta+1}$, we use here that $\omega_1$ is a limit ordinal). It doesn't have a finite subcover (again, limit ordinal) and therefore $\omega_1$ isn't compact.
As for not separable: if it were separable there would be a countable dense subset. But by using the fact that a countable union of countable sets is countable, that's not possible (any countable subset would have an upper bound, and that upper bound would be a countable ordinal $\alpha$; then $\alpha+1$ isn't in the closure of the set).
