Proving that $[0, 1]$ is connected I'm just learning about connectedness and want to show that the interval $[0, 1]$ is connected. However,
I'm very unsure whether my attempt is correct (doubt it).
Attempt:
Suppose there existed disjoint open sets $A, B \neq [0, 1], \varnothing $ s.t. $A \sqcup  B = [0, 1]$, meaning
$A, B$ are open as well as closed. Then one of the values $\sup_{}(A), \inf_{} (A), 
\sup_{}(B), \inf_{} (B) $ is not in $\{0, 1\}$, as one can quickly see. W.l.o.g. let this be
$i_{B} = \inf_{} (B)$. Since $B$ is closed, it must be that $i_{B} \in B$ and further there exists some
$\epsilon>0$ s.t. $(i_{B} - \epsilon, i_{B} + \epsilon)\subseteq B$, in particular $i_{B} - \epsilon / 2 \in B$,
which contradicts the choice of $i_{B}$.
Is this proof sound?
 A: Your proof is correct, but perhaps a bit short concerning the fact $\exists x \in\{\inf(A),\sup(A),\inf(B),\sup(B)\}\colon x  \not\in \{0,1\}$. I think a more transparent approach is to prove that one of $\inf(A),\inf(B)$ must be contained in $(0,1)$. This can be done as follows:

*

*We have $\inf(A) \in A$ and $\inf(B) \in B$ because $A,B$ are closed.


*We must have $\inf(A) < 1$ and $\inf(B) < 1$. Otherwise one of $A, B$ would be the one-point set $\{1\}$ which is not open in $[0,1]$.


*If $\inf(A) > 0$, we are done because $\inf(A) < 1$.


*If $\inf(A) = 0$, it is impossible that $\inf(B) = 0$ because then $0 \in A \cap B$. Thus $\inf(B) > 0$ and we done again as in 3.
So you may assume w.l.o.g. that $\inf(A) \in (0,1)$ and continue as in your qusetion.
Update:
Here is an even shorter proof.

*

*We have $\inf(A) \in A$ and $\inf(B) \in B$ because $A,B$ are closed.


*We must have $\inf(A) = 0$: Suppose that $\inf(A) > 0$. Then, because $A$ is open in $[0,1]$ and $\inf(A) \in A$, there exists an element of $A$ which is smaller than $\inf(A)$. This is a contradiction.


*Similarly we must have $\inf(B) = 0$.


*We get $0 \in A \cap B$, a contradiction.
