Prove that $[0,1]$ is a connected space. Background
I know that there are so many proofs on this that can be found easily on the internet, but I am not familiar with limit points and so on, so I tried to prove it (getting help from a book for the main ideas) using only the basic definitions of open sets, with respect to the topology with open balls.
Below you can find my attempt to prove, and as indicated in the paper. I have 3 questions:


*

*What if $x\gt z-\epsilon$?

*What if $z+\epsilon/2 \gt1$?

*What if $z-\epsilon/2 \lt 0$?


Of course, I know that the idea is that $\epsilon$ is a small number, but how can I formally show that the inequalities that I listed as $1,2,3$ do not hold? Lastly, what is the contradiction in Case 2? I do not really get it.
My Attempt



 A: More simply, show that if $A, B$ are non-empty disjoint open subsets of the space $[0,1],$ with $0\in A,$ then $A\cup B\ne [0,1],$ as follows:
There exists $r\in (0,1]$ such that $[0,r)\subseteq A,$ so $[0,r/2]\subseteq A.$ So the set $$C=\{s\in (0,1]: [0,s]\subseteq A\}$$ is a non-empty subset of $(0,1].$ So $$d=\sup C$$ exists. And $d\in (0,1].$ We claim that $d\not\in A\cup B.$ Proof of claim:
(i).$\; [0,d)\subseteq A.$ Because by def'n of $d=\sup C,$ we have $$x\in [0,d)\implies \exists s\in (x,d)\cap C\implies$$ $$ \implies (\exists s\in (x,d)\cap C \land x\in [0,s]\subseteq A)$$ $$\implies x\in A.$$ (ii). $\;d<1.$ Otherwise by (i), $B=\emptyset$ or  $B=\{1\},$ contrary to $B$ being open and non-empty.
(iii). $\;d\not\in A.$ Otherwise there exists $t>0$ such that $A\supseteq (d-t,d+t)\cap [0,1]$ but then by (i) and (ii) we would have $A\supseteq [0,d)\cup (\, (d-t,d+t)\cap [0,1]\,)\supseteq [\,0, \min (1, d+(t/2)\,)\,],$ implying (by def'n of $C$) that $d=\sup C\ge \min (1, d+(t/2)\,)>d,$ which is absurd.
(iv). $\;d\not \in B.$ Otherwise there exists $u>0$ with $(d-u,d+u)\cap [0,1]\subseteq B.$ But $d>0$ so if $d\in B$ then by (i) we would have $A\cap B\supseteq [0,d)\cap (d-u,d+u)\cap [0,1]\supseteq (\max (0,d-u), d) \ne\emptyset,$ contrary to the hypothesis that $A,B$ are disjoint.
