What shown below is a reference from "Elementos de Topología General" by Fidel Cassarubias Segura and Ángel Tamariz Mascarúa
Definition
A subset $A$ of $\Bbb{R}$ is convex if for any $a,b\in A$ and for any $x\in\Bbb{R}$ such that $a\le x\le b$ it results that $x\in A$.
Proposition
The connected subsets of $\Bbb{R}$ are convex.
Proof. If $Y\subseteq\Bbb{R}$ has more than one point and it is not an interval then there must exist $a,b\in Y$ and $c\notin Y$ such that $a<c<b$ and so the sets $A:=Y\cap(-\infty,c)$ and $B:=Y\cap(c,+\infty)$ are two not empty
disjoint open sets in $Y$ such that $A\cup B=Y$ and so $Y$ is disconnected.So now we suppose that $Y$ is an interval and even we suppose that $Y$ is disconnected. So there must exsist two not empty disjoint open sets $A$ and $B$ of $Y$ such that $A\cup B=Y$. Then without loss of generality we suppose that there exist $a\in A$ and $b\in B$ such that $a<b$ and so we consider $\alpha:=\text{sup}\{x\in\Bbb{R}:[a,x)\cap Y\subseteq A\}$. Clearly it results that $\alpha\le b$ and so, since $Y$ is an interval, $\alpha\in Y$; then by its definition it results that $\alpha\in\text{cl}_Y(A)$ and so, since $A$ is a closed set in $Y$, it results that $\alpha\in A$. Since $A$ is open in $Y$ and since $Y$ is an interval and since $b\in Y\setminus A$ and since $\alpha<b$, there must exist $r>0$ such that $(\alpha-r,\alpha+r)\cap Y\subseteq A$ and so $[a,\alpha+r)\cap Y\subseteq A$, that however is inconsistent with the definition of $\alpha$ and so $Y$ must be connected.
Here for sake of completeness the original text of the proof: I hope mine was a good traslation.
So I don't understand why $\alpha\in\text{cl}_Y(A)$ and why $[a,\alpha+r)\cap Y\subseteq A$. Then I don't understand why if $A$ is open in $Y$ and if $Y$ is an interval and if $b\in Y\setminus A$ and if $\alpha<b$, there must exist $r>0$ such that $(\alpha-r,\alpha+r)\cap Y\subseteq A$: indeed I only see that since $\alpha\in A$ and since $A$ is open in $Y$ it results that $A$ is a neighborhood of $\alpha$ in $Y$ and so there must exist a basic neighborhood $(\alpha-r,\alpha+r)$ in $\Bbb{R}$ of $\alpha$ such that $(\alpha-r,\alpha+r)\cap Y\subseteq A$.
So could someone explain to me what I don't understand, please?
Best Answer
Let $S=\{x\in\mathbb R\mid[a,x)\cap Y\subset A\}$. Then $\alpha=\sup S$. If $x\in S$, the $[a,x)\subset A$, and therefore there are elements of $A$ arbitrarily close to $x$. Now, take $\varepsilon>0$. Then there is some element $s$ of $S$ in $\left(\alpha-\frac\varepsilon2,\alpha\right]$ and, by the definition of $S$, there is some $a\in A$ such that $a\in\left(s-\frac\varepsilon 2,s\right]$. So, $a\in(s-\varepsilon,s]$ and, in particular, $a\in(s-\varepsilon,s+\varepsilon)$. Since every such interval contains an element of $a$, $s\in\overline A$
I will not answer your other question, since you have provided an answer yourself.