I was reading an alternative proof for the connectedness of the unit interval $I = \mathopen[ 0; 1 \mathclose] \subseteq \mathbb R$ in the Euclidean Topology. I managed to follow every step of the proof but the last. I have summarized the main steps, without any justification, to avoid a too lengthy reading. The proof goes as follows.
Since $I$ is connected if and only if the only clopen subsets of $I$ are the empty set and $I$ itself, we start by supposing there is a non-empty clopen set $V \subseteq I $ and then proves that the only possibility is that $V = I$. To do so, it is first proven that $\inf V = 0$. The same technique could be used to prove that $\sup V = 1$, but this would not imply that $V = I$. Instead, the book considers a new set, $$W = \{y \in \mathbb R \mid\mathopen[ 0 ; y \mathclose] \subseteq V \}$$ and then lets $m = \sup W$. We first observe that $m$ is an accumulation point for $W$; then with a few logical manipulations, we imply that $m$ must be also an accumulation point for $V$. Moreover, since by hypothesis $V$ is also a closed set, $V$ must contain its limit points; in particular, it is true that $m \in V$.
Hitherto, everything is clear to me. I have trouble understanding only the last step: namely, the book simply says that
Lastly, the only possibility is that $m = 1$.
I do not see why this is the case. I managed to prove that $m > 1$ leads to a contradiction, but this is trivial. I guess I would have to prove that $m < 1$ too leads to a contradiction, but I do not understand what is holding me from choosing a number $y$ between $m<y<1$, such that $\mathopen [0;y \mathclose ] \nsubseteq V$.
Suppose that $m<1$. Since $V$ is an open set and since $m\in V$, there is some $r>0$ such that $(m-r,m+r)\cap I\subset V$. So, take some $y\in(m,m+r)\cap I$, and then $[0,y]\subset V$, which is impossible, since $y>m=\sup W$.