I’m going to use the more standard notation $\omega_1$ for the first uncountable ordinal.
The most important thing to understand is that the linear order $\le$ on the set $[0,\omega_1]$ is a well-order: if $A$ is any non-empty subset of $[0,\omega_1]$, $A$ has a smallest element with respect to $\le$. That is, there is some $\alpha\in A$ such that $\alpha\le\beta$ for every $\beta\in A$. I’ll use this fact in a bit, but first let’s look a little closer at the elements of $[0,\omega_1]$.
The first countably infinitely many elements are the finite ordinals; you can think of these as being simply the non-negative integers, $0,1,2,3,\ldots$; this is, so to speak, the low end of the order $\le$. Now let $A=\{\alpha\in[0,\omega_1]:\alpha\text{ is not a finite ordinal}\}$. The set of finite ordinals is countable, and $[0,\omega_1]$ is uncountable, so $A\ne\varnothing$, and therefore $A$ has a least (or smallest) element; we call this element $\omega$. The set $\{0,1,2,\dots,\}\cup\{\omega\}$ is still countable, so the set $$[0,\omega_1]\setminus\Big(\{0,1,2,\dots,\}\cup\{\omega\}\Big)$$ is non-empty and therefore has a least element; we call this element $\omega+1$. This $\omega+1$ is the smallest ordinal after $\omega$: it comes right after $\omega$ in the order, so it’s the successor of $\omega$, just as $2$ is the successor of $1$. At this point we have a low end of $[0,\omega_1]$ that looks like this:
$$0,1,2,3,\dots,\omega,\omega+1$$
It should be intuitively clear that we can repeat this argument countably infinitely many times to produce $\omega+2,\omega+3,\dots\,$, and indeed $\omega+n$ for every finite ordinal $n$. Now we have an initial segment of $[0,\omega_1]$ that looks like this:
$$0,1,2,3,\dots,\omega,\omega+1,\omega+2,\omega+3,\dots$$
The only ordinals in this set that are not successors are $0$, since there’s nothing before it at all, and $\omega$, since there is nothing immediately before it: no matter what finite ordinal $n$ you consider, $n+1\ne\omega$.
But this set is still countable, so there is a smallest ordinal in
$$[0,\omega_1]\setminus\{0,1,2,\dots,\omega,\omega+1,\omega+2\dots\}\;;$$
this ordinal is denoted by $\omega\cdot 2$, and like $\omega$, it’s not a successor: it is not $\alpha+1$ for any $\alpha$. In other words, it’s a limit ordinal, as is $\omega$. ($0$ is a bit anomalous: it’s not a successor ordinal, but it’s also not a limit ordinal.)
After this the ordinals and their standard notations start getting a bit complicated, and we don’t really have to go into them to understand the topological properties of $[0,\omega_1)$ and $[0,\omega_1]$. It’s also true that the usual set-theoretic definition of the ordinals makes each ordinal equal to the set of its predecessors:
$$1=\{0\},2=\{0,1\},3=\{0,1,2\},\dots,\omega=\{0,1,2,\dots\},\omega+1=\{0,1,2,\dots,\omega\}\,,$$
and so on; this is the boxes within boxes that you mentioned. However, you don’t need to worry about this, either.
Now let’s see why every strictly decreasing sequence in $[0,\omega_1]$ is finite. Suppose that we had an infinite sequence $\langle\alpha_n:n\in\Bbb N\rangle$ such that $\alpha_0>\alpha_1>\alpha_2>\ldots\,$; then the set $A=\{\alpha_n:n\in\Bbb N\}$ would be a non-empty subset of $[0,\omega_1]$ with no least element, contradicting the fact that $[0,\omega_1]$ is well-ordered. Infinite increasing sequences are no problem at all, however: for each $\alpha\in[0,\omega_1)$, the set $[0,\alpha]$ is countable, so $[0,\omega_1)\setminus[0,\alpha]\ne\varnothing$, so there are elements of $[0,\omega_1)$ bigger than $\alpha$. The smallest of these is $\alpha+1$, the successor of $\alpha$. Thus, starting at any $\alpha\in[0,\omega_1$ I can form an infinite increasing sequence $\langle\alpha,\alpha+1,\alpha+2,\dots\rangle$ whose members are all still in $[0,\omega_1)$.
Next, let’s see why $[0,\omega_1)$ is first countable. Let $\alpha\in[0,\omega_1)$. Suppose first that $\alpha$ is a successor ordinal, say $\alpha=\beta+1$; then $(\beta,\alpha+1)=[\beta+1,\alpha+1)=[\alpha,\alpha+1)=\{\alpha\}$ is an open nbhd of $\alpha$ in the order topology, so $\alpha$ is an isolated point, and $\big\{\{\alpha\}\big\}$ is certainly a countable local base at $\alpha$! Note that $0$ behaves like a successor ordinal: $[0,1)=\{0\}$ is an open nbhd of $0$, so $0$ is also an isolated point.
Now suppose that $\alpha$ is a limit ordinal. For each $\beta<\alpha$ the set $(\beta,\alpha+1)=(\beta,\alpha]$ is an open nbhd of $\alpha$. Every open nbhd of $\alpha$ contains an open interval around $\alpha$, which in turn contains one of these intervals $(\beta,\alpha]$, so $$\mathscr{B}_\alpha=\Big\{(\beta,\alpha]:\beta<\alpha\Big\}$$ is a local base at $\alpha$. Finally, $\alpha<\omega_1$, and $\omega_1$ is the first ordinal with uncountably many predecessors, so there are only countably many $\beta<\alpha$, and $\mathscr{B}_\alpha$ is therefore countable. Thus, every point of $[0,\omega_1)$ has a countable local base, and $[0,\omega_1)$ is therefore first countable.
Note that $[0,\omega_1]$ is not first countable, because there is no countable local base at $\omega_1$: if $\big\{(\alpha_n,\omega_1]:n\in\Bbb N\big\}$ is any countable family of open intervals containing $\omega_1$, let $A=\bigcup_{n\in\Bbb N}[0,\alpha_n]$. Then $A$, being the union of countably many countable sets, is a countable subset of $[0,\omega_1)$, so $[0,\omega_1)\setminus A\ne\varnothing$. Pick any $\beta\in[0,\omega_1)\setminus A$; then $(\beta,\omega_1]$ is an open nbhd of $\omega_1$ that does not contain any of the sets $(\alpha_n,\omega_1]$, and therefore the family $\big\{(\alpha_n,\omega_1]:n\in\Bbb N\big\}$ is not a local base at $\omega_1$. That is, no countable family is a local base at $\omega_1$, so $[0,\omega_1]$ is not first countable at $\omega_1$.
Finally, let’s look at compactness. Suppose that $\mathscr{U}$ is an open cover of $[0,\omega_1]$. Then there is some $U_0\in\mathscr{U}$ such that $\omega_1\in U_0$. This $U_0$ must contain a basic open nbhd of $\omega_1$, so there must be an $\alpha_1<\omega_1$ such that $(\alpha_1,\omega_1]\subseteq U_0$. $\mathscr{U}$ covers $[0,\omega_1]$, so there is some $U_1\in\mathscr{U}$ such that $\alpha_1\in U_1$. This $U_1$ must contain a basic open nbhd of $\alpha_1$, so there is some $\alpha_2<\alpha_1$ such that $(\alpha_2,\alpha_1]\subseteq U_2$. Continuing in this fashion, we can construct a decreasing sequence $\alpha_1>\alpha_2>\alpha_3>\ldots\,$, which, as we saw before, must be finite. Thus, there must be some $n\in\Bbb Z^+$ such that $\alpha_n=0$, and at that point $\{U_0,\dots,U_n\}$ is a finite subcover of $\mathscr{U}$.
The space $[0,\omega_1)$, on the other hand, is not compact. It is countably compact, however. The easiest way to prove this is to show that $[0,\omega_1)$ has no infinite, closed, discrete subset. Suppose that $A$ is a countably infinite subset of $[0,\omega_1)$. Let $\beta$ be the smallest element of $[0,\omega_1)$ that is bigger than infinitely many elements of $A$. (You’ll have to explain why $\beta$ exists, using the fact that $A$ is countable and $[0,\omega_1)$ is well-ordered.) Finally, show that $\beta$ is a limit point of $A$. Then either $\beta\notin A$, in which case $A$ isn’t closed, or $\beta\in A$, in which case $A$ isn’t discrete.
