Closed unit interval is connected proof

$A$ meets $\mathbb{I}$ non-trivially so $c$ is bigger than $0$ necessarily. Since $b<1$ and $b$ is an upper bound for $\{ x \in A : x < b \} $, $c \leq b < 1$. So, $0<c<1$.

It follows that $c \in (0,1) \subseteq [0,1] = A \cup B$ by supposition.

Now we'll assume that $c \in A$.

Necessarily $c < b$.

If $c=b$ then $c \in A \cap B$ however $A \cap B = \varnothing$ and also $c$ cannot exceed $b$.

We can pick some $\varepsilon_1$ neighborhood around $c$ that is contained in $A$ since it is an open set. Pick some $c< a_1< c+\varepsilon_1$ in that neighborhood. $A$ and $B$ are disjoint so this neighborhood will not contain $b$ so this $a_1 < c+\varepsilon_1 < b$ so $a_1 \in \{ x \in A : x < b \}$ but then $a_1 \leq c$ since it is the supremum but also we have $a_1 > c$, a contradiction.