Is there a way to show the following: for any two numbers in [n]={1,2,...,n}, there is a permutation that sends them into two consecutive numbers in [n].
i.e, Choose $i,j\in [n]$, for any k, with $1\leq k\leq n-1$, there is a permutation $\sigma_{_{k}}$ such that $i\ \sigma_{_{k}} = k$ and $j\ \sigma_{_{k}}= k+1$. I think the permutation depends on the position of $i,j$. I mean if $i,j$ are consecutaive then the identity is our desire permutation. However, if $i,j$ are not consecutive then the permutation is the one that sends $i\mapsto j-1,$ and $j\mapsto j$.
i.e, $2,4\in [5]$ and the permutation is for example $(23)$ or $(123)$. the problem that I couldn't prove it.
A permutation of the set $S = \{1,2,\ldots,n\}$ is just a rearrangement of the elements of $S$. The statement that you are trying to prove is almost trivially true.
For instance, choose $i,j \in S$. Then for $k$ such that $1\leq k < n$ let $\sigma_k$ be the permutation that swaps $i$ and $k$, swaps $j$ and $k+1$, and leaves all other elements of $S$ fixed. This is one explicit example of a permutation with the property you desire, but there are many more. Can you see why?