I wrote a proof, and I just wanted to verify if it is correct.
Proof : Suppose not. Then, $[0,1] = U\cup V$ for $U$ and $V$ open in $[0,1]$ such that $U\cap V = \emptyset$. Let $0\in U$. Consider $s= sup\{t \mid [0,t] \subset U\}$. Clearly, $s\in[0,1]$.
If $s\in U$, then since $U$ is open, $\exists \epsilon >0$ such that $B(s;\epsilon) \subset U$ . But then $s+\epsilon \in U$ , and $s+\epsilon > s$, contradicting the definition of $s$. So, $s \notin U$.
Then, $s\in V$. But again, since $V$ is open, $\exists \epsilon >0$ such that $B(s;\epsilon)\subset V$. But, since $s$ is the least upper bound of $U$, for this $\epsilon$, $\exists s_{1} \in U$ such that $s_{1} \in B(s;\epsilon)$. Then, $s_{1} \in U\cap V$, which contradicts the fact that $U$ and $V$ are disjoint.
So, our assumption is wrong and $[0,1]$ is connected.
Best Answer
You're probably not looking for the answer still after a year, but for those who come across this in the future, it may be helpful. You are pretty much right about your proof, but here is a more formal way of writing it.
That is that "If $a<b$, then the subspace $[a,b]$ of $E^1$ is connected".
Let $a<b$ and let $[a,b]$ be a subspace of $R$ with the $E^{1}$ topology. For the sake of contradiction, assume that $[a,b]$ is not connected. Then $[a,b]= U \cup V$, where $U$ and $V$ are nonempty disjoint open sets in $[a,b]$. Without loss of generality we may assume that $b \in V$. Because $U$ is nonempty and bounded above (by $b$), the axiom of completeness states that $U$ has a least upper bound $s$. We will prove that $s$ is not an element of either $U$ or $V$, and this will yield a contradiction.
Suppose $s \in U$. Since $U$ is open, there exists an $\epsilon> 0$ such that $B_{\epsilon}(s) \subseteq U$. Thus, since $s+ \frac{\epsilon}{2} \in U$ and $s<s+ \frac{\epsilon}{2}$, there exists an element in $U$ that is greater than $s$. Therefore, $s$ is not an upper bound for $U$. This is a contradiction. Thus, $s \not \in U$. Now suppose $s \in V$. Then, because $V$ is open, there exists an $\epsilon>0$ such that $B_{\epsilon}(s) \subseteq V$. Then $s- \frac{\epsilon}{2}$ is an upper bound for $U$. But $s- \frac{\epsilon}{2}<s$, which is a contradiction because $s$ is the least upper bound for $U$. So $s \not \in V$. Thus, $s \not \in U \cup V$, which is a contradiction. Therefore, $[a,b]$ is connected.