Explanation of proof that intervals are connected subspaces of $\Bbb R$ Maybe it's the late hours, but I've hit a brick wall trying to parse this seemingly innocuous proof that intervals are connected and would really appreciate some help. The proof goes as follows:

Let $I$ be an interval (a set that satisfies the property $x,y\in I\implies[x,y]\subset I$) and assume it is not connected. Then there exist $O_1,O_2\subset I$ disjoint, non-empty and open in $I$, such that $I=O_1\cup O_2$. Let $x_i\in O_i$ and w.l.o.g. assume $x_1<x_2$. Let $s=\sup\{y\in O_1:y\leq x_2\}$ (the supremum exists because the set is non-empty and bounded from above). Since the $O_i$ are open in $I$, there exists an $\varepsilon>0$ such that $(x_i-\varepsilon,x_i+\varepsilon)\cap I\subset O_i$ (follows from the definition of the subspace topology).

Past this point I have trouble following:

From this it follows that $x_1+\varepsilon\leq s\leq x_2-\varepsilon$ (why?) and that $s\in(x_1,x_2)$ and therefore $s\in I$. It follows that $s$ belongs to either $O_1$ or $O_2$ and because of the fact that they're open (shouldn't it be "open in $I$"?) there exists a $\delta>0$ such that $(s-\delta,s+\delta)\subset O_i$ (why? we just know that $O_i$ is open in $I$, all we can conclude by the definition of the subspace topology is that there exists a $\delta$ such that $(s-\delta,s+\delta)\cap I\subset O_i$). Either case contradicts the definition of $s$, since if $(s-\delta,s+\delta)\subset O_1$ then $s<s+\delta\leq\sup\{y\in O_1:y\leq x_2\}$ and if $(s-\delta,s+\delta)\subset O_2$ then $\sup\{y\in O_1:y\leq x_2\}\leq s-\delta<s$ (why? I can see why the first inequality follows, but not why the second and again, I don't understand why an entire ball around $s$ has to be included along with it).

 A: Let me fill in some of what the author did not state explicitly.
Let $O_i = I \cap U_i$ for $U_i$ open in $\mathbb{R}$
We know that there exists $\varepsilon_i>0$ such that $(x_i - \varepsilon_i, x_i + \varepsilon_i) \subset U_i$, since $U_i$ is open.
Since we can find these two $\varepsilon_i$, there also exists $\varepsilon>0$ such that $(x_i - \varepsilon, x_i + \varepsilon) \subset U_i$. (i.e. one constant for both sets)
We are given that $x_1<x_2$.
If $x_2 < x_1 + \varepsilon$, then $x_2 \in (x_1 - \varepsilon, x_1 + \varepsilon) \subset U_1$.
We know that $x_2 \in O_2 \subset I$.
Combining these gives that $x_2 \in O_1$, which contradicts that $O_1 \cap O_2 = \emptyset$.
Thus, we have that $x_2 \ge x_1 + \varepsilon$.
Since $I$ is an interval, we know that $[x_1,x_2] \subset I$, so that $(x_1, x_1 + \varepsilon) \subset I$.
Hence $(x_1, x_1 + \varepsilon) \subset U_1$ and $(x_1, x_1 + \varepsilon) \subset I$, so $(x_1, x_1 + \varepsilon) \subset O_1$.
The symmetric argument would show that $(x_2 - \varepsilon, x_2) \subset O_2$
Combining those both gives that $x_1 + \varepsilon \le s \le x_2 - \varepsilon$
Now, $s \in [x_1, x_2] \subset I$ means that either $s\in O_1 \subset U_1$ or $s\in O_2 \subset U_2$.
Either way, there exists $\delta > 0$ so that either $(s - \delta, s + \delta) \subset U_1$ or $(s - \delta, s + \delta) \subset U_2$
Let's first consider $(s - \delta, s + \delta) \subset U_1$
Using the same argument as before, we can see that if $x_2 < s + \delta$, then $x_2 \in (U_1 \cap I)$, which contradicts $O_1 \cap O_2 = \emptyset$, so $x_2 \ge s + \delta$, and hence $(s, s + \delta) \subset [s, x_2] \subset I$, so that $(s, s + \delta) \subset O_1$.
Now we have our contradiction, because $s + \delta/2 \in O_1$ and $s + \delta/2 \le x_2$, but $s + \delta/2 > s$, which contradicts the definition of $s$ in that $s$ is not an upper bound.
For the case $(s - \delta, s + \delta) \subset U_2$, by a symmetric argument we get that $(s - \delta, s) \subset O_2$, and that $s - \delta/2$ is an upper bound of $\{y \in O_1: y\le x_2\}$, which contradicts that $s$ is the least upper bound.
