Question about proof that $I \subset \mathbb{R}$ is a connected set. Let $I \subset \mathbb{R}$ be an interval or all of $\mathbb{R}$, then $I$ is a connected set.
Proof. Suppose $I$ is an interval but not connected, then we would have $I \subset U \cup V$ where $U$ and $V$ are both non-empty and open, $U \cap I \ne \emptyset$, $V \cap I \ne \emptyset$, and $U \cap V = \emptyset$. Let $a \in U$ and $b \in V$. Without loss of generality, we can assume that $a < b$ (otherwise just change the names of $U$ and $V$). Let $U_{1} = U \cap I \cap [a,b]$ and $V_{1}=V \cap I \cap [a,b]$. Then $U_{1}$ and $V_{1}$ are disjoint, non-empty sets because $a \in U_{1}$ and $b \in V_{1}$. Since $U_{1}$ is bounded, $c = \sup\{x: x \in U_{1}\} < \infty$. From $b \in V$, $a <b$, and $V$ being open, we know there is $\delta >0$ such that $(v-\delta, v) \subset V_{1}$. Let $\delta$ be the maximal choice that satisfies this inclusion in $V_{1}$ (this exists because we know $\delta \le b-a$). We then know $a \le c \le b-\delta$. Moreover, by the assumption on $\delta$, $c \notin V$, otherwise the openness of $V$ would allow us to increase $\delta$. Since $c$ is not the right limit of $I$, we also know that $c \notin U$, otherwise by the openness of $U$ and $c<b$, it would not be the supremum of $U_{1}$ 
I was hoping that someone could explain the following highlighted sentence.

Since $c$ is not the right limit of $I$, we also know that $c \notin U$, otherwise by the openness of $U$ and $c < b$, it would not be the supremum of $U_{1}$.

How do we know that $c$ is not the right limit of $I$? I also don't understand how we know that $c \notin U$, otherwise by the openness of $U$ and $c <b$, it would not be the supremum of $U_{1}$.
 A: $c$ cannot be the right limit of $I$ because $c<b$ and $b\in I$.
$c\notin U$ because $c$ is the supremum of $U$, which is open.  In other words, if $c\in U$, then some neighborhood of $c$ would be included in $U$, but then there would have to be a value greater than $c$ in $U$, which would mean that $c$ wasn't the supremum.
A: Maybe it's better to streamline this. You are trying to prove that $c:=\sup U_1$ is neither in $U_1$ nor $V_1$, which gives us an immediate contradiction because $c\in [a,b]$ and $[a,b]\subseteq U_1\cup V_1.$
If $c\in V_1$ then $c \neq a$, so either $c = b$ or $a < c < b$. But $V_1$ is open, so it contains an interval $(d,c]$. If $c=b,$ we are done because now $d$ is a smaller upper bound on $U_1$ than $c$, which is a contradiction. And if $c<b,$ then since $c$ is assumed to be an upper bound on $U_1$ the interval $(c,b]$ is contained entirely in $V_1$, so in fact $(d,b]$ is entirely contained in $V_1$ and we get a contradiction because now again $d$ is a smaller upper bound on $U_1$ than $c$.
If $c\in U_1$, the argument is even easier, so I will leave it to you. 
Remark: when I do these kinds of proofs, I always draw pictures first, to see what is going on.
A: Perhaps simpler to let $d=\{x\in [a,b]: [a,x)\subset U\}.$ There exists $r>0$ such that  $[a,a+r)\subset (a-r,a+r)\subset U,$ so  $a<\sup d\le b.$
Consider $[a,\sup d).$
For any $x\in [a,\sup d),$  by def'n of $\sup$ there exists $y\in d$ with $x\le y\le \sup  d.$ And $y\in d\implies [a,y)\subset U,$ so $[a,x)\subset [a,y)\subset U.$ So $[a,x)\subset U$ for any $x\in [a,\sup d).$ 
Hence $[a,\sup d)=\cup \{[a,x):x\in [a,\sup d)\}\subset U.$
So $[a,\sup d)\subset U.$
Now $\sup d\in V$ because
(i) if $b=\sup d$ we already have $b\in V.$
(ii) if $b>\sup d$ and $\sup d \in U$ then there exists $s\in (0,b-d)$ such that $(\sup d -s,\sup d+s)\subset U,$ implying $[a,\sup d+s)=[a,\sup d)\cup [\sup d,\sup d+s)\subset U.$  By def'n of $d$ this makes $\sup d+s\in d,$ so $\sup d+s$ is a member of $d$ which is greater then  $\sup d ,$ which is absurd. So $b>\sup d\in U$ is untenable, so $b>\sup d\implies \sup d \in V.$
Now since $\sup d\in V$ there exists $t>0$ with $(\sup d -t,\sup d+t)\subset V.$  Recall that $\sup d>a$  and $[a,\sup d)\subset U.$ So we have    $$\emptyset \ne [a,\sup d)\cap (\sup d -t,\sup  d)\subset U\cap V\cap [a,b] \subset (U\cap I)\cap (V\cap I)$$ contrary to the hypothesis that $(U\cap I)$ and $(V\cap I)$ are disjoint.
