Every interval $I \subset \mathbb{R}$ is connected. [Proof clarification]

The first two bolded statements are less about using ideas from abstract topological spaces and more about using the definition of infimum, properties of real intervals and manipulating inequalities:

In what follows let us denote $B_a = \{r \in B\ :\ a < r\}$.

Then in every neighborhood of $s$ there are points of $B$ (because of the definition of the infimum), ...

Suppose $(p, q)$ is a basic neighborhood of $s$ in $\mathbb{R}$; so $ p < s < q$. Now since $s$ is the largest lower bound of $B_a$ and $q$ is larger than $s$, $q$ cannot be a lower bound of $B_a$.

So there must be a point $x$ of $B_a$ (and thus of $B$) such that $x < q$. Also since $s \leq x$ by the lower bound property of $s$, we have $p < s \leq x < q$ i.e. $x \in (p, q)$. Thus every basic neighborhood $(p, q) \cap I$ of $s$ in $I$ with the subspace topology, still contains that point $x$ of $B$ and we are done.

... but also of $A$, then if not $s = a$, then $a < s$ and the open interval $(a,s)$ lies entirely in $A$.

Again, consider our neighborhood $(p, q)$ of $s$ from earlier. If $s = a$, we are done because then $s$ is a common element of $A$ and the neighborhood $(p, q) \cap I$ of $s$ as required.

Otherwise, if $s \neq a$, then $a < s$. Again this is because of the largest lower bound property of $s$. $a$ is a lower bound on $B_a$ by definition. So since $s$ is the largest lower bound of $B_a$, $a \leq s$. Along with $s \neq a$ this implies $a < s$.

Now analyze the number $$x = \frac{\max(a, p) + s}{2}$$ Since both $p < s$ and $a < s$ in this case, we have $a, p \leq \max(a, p) < s$. So $p < x < s$ which implies $x \in (p, q)$. And $a < x < s \leq b$ which implies $x \in I$ because $I$ is an interval. Finally as $s$ is a lower bound on $B_a = \{r \in B\ :\ a < r\}$, any real $r$ which is less than $s$ but still $a < r$ cannot be in $B$. So $a < x < s$ implies $x \not\in B$. And this gives $x \in A$ as $A$ and $B$ partition $I$.

... And so $s$ cannot be an inner point of $A$ nor $B$, ...

I understand why this compact statement might lead you to think of $s$ being a boundary point of either $A$ or $B$. That is certainly a correct intuition which can be used to complete the proof. But the direct intention of this statement was much simpler: it was intending to derive a contradiction by showing that either $A$ is not open or $B$ is not open like so:

Note that $a \leq s \leq b$ which gives $s \in I = A \cup B$ as $I$ is an interval. Now:

• if $s \in A$, then by definition of $A$ being open, there is a basic neighborhood $U = (p, q) \cap I$ of $s$ entirely contained in $A$. So since $A$ is disjoint from $B$, $U$ cannot contain any point of $B$. But this contradicts what we just showed above: every such $U$ contains a point of $A$ and a point of $B$.

• and if $s \in B$, we get a completely analogous contradiction.

1st question. Let D = { x in B : a < x }. Let U be a nhood of s = inf D. Thus some u,v with s in (u,v) subset U. If not some x in D with x < v, then for all x in D, v <= x.
As that contradicts s = inf D, exists x in D with s <= x < v.
Consequently that x which is in B, is in (u,v), thus in U.

2nd question using notation from above.
If U $\cap$ A is empty, then for all x in A, x <= u or v <= x.
Thus (u,v) subset B.
As a <= u, exists x in (u,s) $\cap$ D, a contradiction.
Thus U $\cap$ A is not empty.

Finally, either s in A or s in B.
If s in A, then as A is a nhood of s, points of B are in A, a no no.
If s in B, then as B is a nhood of s, points of A are in B, a no no.