Proof that $(0,1)$ is connected. I wasn't completely satisfied with the proof given by my professor that the open interval $(0,1)$ is connected in the subspace topology of Euclidian space. So I came up with this argument instead, please tell me if there is anything unclear about this proof. Am I using any unexplicit assumptions?
A proof by contradiction, assume that $(0,1)$ is disconnected. Let $A,B$ be disjoint non-empty open sets such that $X=A\bigcup B$. Then $A,B$ are also closed in $(0,1)$. Since $A=\{x\in (0,1) :x\in A\}$ we have that A is bounded by $1$, since $A$ is non-empty by assumption, the $LUB$ axiom tells us that $sup(A)=s$ exists. Furthermore $s\in \bar{A}$, since for all $r> 0$ there exists $x\in A$ such that $s-r<x<s$ because othervise $s-r$ would be the supremum of $A$. Hence $B_r(s)\bigcap A\neq \emptyset$, this shows that $s\in \bar{A}$. Since $A$ is closed $A=\bar{A}$ and hence $s\in A$. Since $A$ is open $B_{\epsilon }(s)\subseteq A$, i.e. $(s- \epsilon , s + \epsilon ) \subset A$, this implies that $s + \frac{\epsilon }{2} \in A$, contradicting the assumption that $s=sup(A)$.
 A: Your proof that $s\in\operatorname{cl}A$ is slightly flawed. At that point in the argument you don’t know that for each $r>0$ there is an $x\in A\cap(s-r,s)$; you know only that there is an $x\in A\cap(s-r,s]$. That is, $s$ could in principle be the supremum of $A$ not because it’s a limit from the left of elements of $A$, but simply because it’s the maximum element of $A$. An example would be the set $\left[0,\frac12\right)\cup\left\{\frac34\right\}$. Of course we know that an open set $A$ can’t really look like this, but you’d have to make a separate argument to show that $s$ is not just $\sup A$, but actually $\sup(A\setminus\{s\})$.
However, this flaw doesn’t affect the conclusion that $s\in\operatorname{cl}A$: if $s\in A$, then automatically $s\in\operatorname{cl}A$, and if $s\notin A$, then your argument works.
The other problem is that $s$ might be $1$. In that case you could replace $A$ with $B$ and go for the same basic argument, but what if $\sup A=\sup B=1$? To finish your proof, you’d have to show that that’s impossible, but I think that you’ll find that a bit tricky.
