Show that $[0,1]$ is connected in the usual topology $(\Bbb R, \tau)$. Show that $D=[0,1]$ is connected in the usual topology $(\Bbb R,\tau)$.
EDIT:
Attempt:
For the sake of contradiction, suppose that $D$ is disconnected. Then there exists two open sets $A$ and $B$ for which
$A \cap D \ne \emptyset, B \cap D \ne \emptyset, (A \cap D) \cap (B \cap D) = \emptyset$, and $(A \cap D) \cup (B \cap D) = D$. Now, since $A \cap D \ne \emptyset$, there is $a \in A \cap D$ such that $a>0$. Similarly, since $B \cap D \ne \emptyset$, there is $b \in B \cap D$ such that $b<1$. Since $(A \cap D) \cap (B \cap D)=\emptyset$, then $a \ne b$. Without loss of generality, let $a<b$. Then $0<a<b<1$. Next, define a set $E=\{x \in A : x<1\}$.
Since $a \in E$, then $E$ is exists.
Since for any $x \in E, x<1$, then $1$ is an upper bound of $E$, so $E$ has a supremum, say $c$. Then
\begin{align*}
c \in D=[0,1]&=(A \cap D) \cup (B \cap D) \\
&= (A \cup B) \cap D.
\end{align*}
There are two cases should be considered, that is $c\in A$ and $c \in B$.

*

*If $c \in A$, since $A$ is open, then there exists an open neighbourhood $U$ of $c$ for which
$$c \in U \subseteq A.$$
It means that there is an open interval
$(a_A,b_A)$ such that $c \in (a_A,b_A) \subseteq U \subseteq A$.
Hence, $a_A < c < b_A$. Now, $c<b_A<c+b_A$. Since $A \cap D$ and $B$ are disjoint, then $c+b_A<b$. Therefore, $b_A<1$ and so $b_A \in E$, which contradicts the definition of $c$ being supremum of $E$.
Thus, $c \not \in A$.


*If $c \in B$, since $B$ is open, then there exists an open neighbourhood $V$ of $c$ for which $c \in V \subseteq B$.
I got stucked at 2. Any ideas?
 A: Using your approach, since $1\in D$, we can assume that $1\in B$. Therefore, since $A\cap B\neq \emptyset$ and $B$ is open, $s=\text{sup} A< 1$. Since $A$ is open, this supremum cannot be a maximum. Hence $s\in B$, but since $B$ is open, there exists some $\epsilon>0$ such that $s-\epsilon \in B$, but because $s$ is the supremum of $A$, we can take this $\epsilon$ small enough such that $s-\epsilon$ is also in $A$. Contradicting the fact that $A \cap B=\emptyset$.
A: Hint: You know that the topology on $[0,1]$ is generated by the open sets
$$\{[0,a), (b, 1], (c, d)~|~a \in (0,1], b\in [0,1), ~c, d \in (0,1)\}.$$
Now try to show that $[0,1]$ cannot be covered by two open sets as above.
A: U don't need to define $E$, $A, B$ are non-empty and bounded, since they are contained in the unit interval, so $\sup(A), \inf(A), \sup(B), \inf(B) $ exists. use the characterization of the supremum $c=sup A \Leftrightarrow \forall x\in A, x\leq c \land \forall \epsilon >0, \exists a_{\epsilon} \in A, a>c-\epsilon $, in particular $c\not \in A$, so it must belong to $B$, but this is impossible also by the characterization above, in fact if $c\in B, \exists \epsilon\ >0, \left] c-\epsilon, c+\epsilon \right [\cap \left [0,1\right] \subset B$, So $a_{\epsilon} \in B\cap A$, but the intersection of $B$ and $A$ is empty.
