Proof for the fact that [0,1] or any interval [a,b] in $\mathbb{R}$ is connected I found a proof for the fact that any interval [a,b] is connected in $\mathbb{R}$ can somebody check if it is correct.
Suppose that if it is not connected then $\exists$ two closed sets U and V such that they are disjoint and non empty and  their union is [a,b] now since they are disjoint and non empty consider distinct points $u_{0}$ in U and $v_{0}$ in V now construct the sequences $\{u_{n}\}$ and $\{v_{n}\}$ as follows $m =\frac{u_{i} +v_{i}}{2}$ if m $\epsilon$ U then $u_{i+1} = m$ , $v_{i+1}=v_i$ else if m belongs to V then $v_{i+1} =m$ , $u_{i+1}=u_{i}$ now we get sequences $\{u_{n}\}$  and $\{v_{n}\}$ both converge to the same value c now since c is a limit point for both U and V implies c belongs to both U and V (bcoz U and V are both closed ). Hence contradiction because we assumed them to be disjoint.
 A: Your proof relies on the metric and some arithmetic on $[0,1]$. Topologists like to go basic and in this case we can use that $[0,1]$ (or the homeomorphic $[a,b], a < b$) have the order topology and are nice:
Some general theory:
Let $X$ be an ordered topological space $(X,<)$, so the topology is generated by the subbase $\{(\leftarrow,x),(x,\rightarrow)\mid x \in X\}$
Definition: A cut $(A,B)$ of $X$ (by which I mean $A,B \subseteq X$, both non-empty, $A \cap B = \emptyset$, $A \cup B = X$, and also for all $a \in B$ and all $b \in B$ we have $a < b$) is called a jump if $A$ has a maximum and $B$ has a minimum, and a gap if neither is the case.
Theorems:

*

*$X$ is connected iff $X$ has no gaps or jumps.

*$X$ is compact iff it has no gaps, and a minimum and a maximum.


$[0,1]$ has no gaps as when $(A,B)$ is a cut of $[0,1]$, $\sup A$ is well-defined and if it lies in $A$, $A$ has a maximum, and if it lies in $B$, $B$ has a minimum. So no gap can exist.
Immediate corollary from the second theorem is that $[0,1]$ is compact (as $0=\min[0,1], 1= \max[0,1]$ also exist).
$[0,1]$ also has no jumps because between a supposed $\max A$ and $\min B$ we always have at least one rational which would be in neither $A$ not $B$ (by construction of $\Bbb R$ from $\Bbb Q$).
So the first theorem then implies $[0,1]$ is connected.
These theorem have a lot more applications than just the intervals in $\Bbb R$, and also apply to lexicographically ordered squares and ordinal spaces etc. So it's quite useful to have these general criteria.
My alma mater had quite a few specialists on ordered spaces so these theorems were standard fare in our topology diet as students. Nowadays not many texts cover them any more, alas.
