Questions about Munkres’ proof that the real line is connected. I was reading Munkres' proof that the real line is connected and I have had a lot of trouble understanding it. Particularly why there are two cases one being that the $\sup A_0$ belongs in $B_0$ and the other being that the $\sup A_0$ belongs in $B_0$.
Also why is $d$ a smaller upper bound than $c$ in $A_0$? And where does the contradiction lead us?
I am really confused about this, so any help would be much appreciated
 A: Aiming to get a contradiction, Munkres starts with a separation $\{A_0,B_0\}$ of the interval $[a,b]$, where $a\in A_0$ and $b\in B_0$. Because we’re working in the linear continuum $L$, we know that $\sup A_0$ exists. Since $A_0\subseteq[a,b]$, we know that $a\le x\le b$ for each $x\in A_0$, so $a\le\sup A_0\le b$. Let $c=\sup A_0$; then $c\in[a,b]=A_0\cup B_0$. Moreover, $A_0\cap B_0=\varnothing$ so either $c\in A_0$, or $c\in B_0$. Munkres will get a contradiction by showing that in fact neither of these is possible: $c\notin A_0$, and $c\notin B_0$; his Case $\mathit{1}$ is the proof that $c\notin B_0$, and his Case $\mathit{2}$ is the proof that $c\notin A_0$. Each of these is itself a proof by contradiction.
The quantity $d$ appears in Case $\mathit{1}$, in which we’re trying to prove that $c\notin B_0$. To get a contradiction, we suppose that $c\in B_0$. In particular this means that $c\ne a$, since $a\in A_0$, so either $c=b$, or $a<c<b$. By hypothesis $B_0$ is open in $[a,b]$, so in either case there a $d<c$ such that $(d,c]\subseteq B_0$.
Now suppose that $x\in A_0$. If $x>d$, then either $x\in(d,c]\subseteq B_0$, which is impossible, or $x>c$, which is also impossible, so in fact $x\le d$. And $x$ was an arbitrary element of $A_0$, so every element of $A_0$ is less than $d$, and therefore $d$ is an upper bound for $A_0$, contradicting the fact that $c$ is the least upper bound of $A_0$. This contradiction shows that there cannot be an element $d\in[a,b]$ such that $d<c$ and $(d,c]\subseteq B_0$. But $B_0$ is open, so if $c\in B_0$, there must be such a $d$, and therefore $c$ cannot be in $B_0$ after all.
In Case $\mathit{2}$ Munkres goes on to show, again by contradiction, that $c$ cannot be in $A_0$, either, and therefore cannot exist, contradicting the fact that it must exist, since $L$ is assumed to have the least upper bound property.
