Why $[0,1]$ is connected I try to prove that $[0,1]$ is connected.I have to  follow  the next steps: $R$ is complete using the construction of $R$ with Cauchy sequences $\Rightarrow$ $[0,1]$ is connected.How can I prove that?
 A: You can prove it using that outline, but as it's not generally true that completeness implies compactness implies connectedness you will need to use more facts here.
The first step is normally left out in early courses. Or one cheats by using some axiom about completeness. For example we used the axiom of least upper bound. This is used together with a Cauchy-sequence $x_n$ by first noting that you can construct two sequences $\inf_{j\ge n} x_j$ and $\sup_{j\ge n}x_j$, the first being increasing and the second decreasing. You then use the axiom again to see that these are convergent and that they converge to the same value which is in fact the same as $x_n$ converges to. We here uses the fact that $[0,1]$ is bounded to be allowed to use the axiom of least upper bound and also it being a closed interval to guarantee that the sequences converges within it.
For compactness you can prove this going via the property that every sequence in $[0,1]$ has a subsequence that's a Cauchy-sequence. You can do this by successively part the interval in two equal parts - at least one of them will contain infinite number of numbers from the sequence. You take this interval and selects the first number in that interval. Then you repeat with that interval, and the sequence following the selected number. Then you get a subsequence that's a Cauchy-sequence and therefore convergent. 
Then you can use compactness to prove connectedness. Assume that you could decompose the interval in to disjoint open sets $A$ and $B$. Now since $A$ and $B$ are open they are themselves union of open intervals and since $[0,1]$ is covered by these intervals they could be reduced to a finite number of intervals covering $[0,1]$. You can then replace overlapping intervals with a larger interval and consequently constructing an disjoint open covering of $A$ and one for $B$ which would cover $[0,1]$. Now order the interval and consider two adjacent (open, non-overlapping) intervals. Now the lower limit of the upper interval and the upper limit of the lower are not part of the covering which contradicts the decomposition. In this part you use again that $[0,1]$ is a closed interval since the limits of the last sets to be within $[0,1]$.
