Show that if n+1 integers are choosen from set $\{1,2,3,...,2n\}$ ,then there are always two which differ by 1

Considering n=5 i have $\{1,2,3,...,10\}$ .Making pairs such as $\{1,2\}$ ,$\{2,3\}$ ... total of $9$ pairs which are my holes and $6$ numbers are to be choosen which are pigeons .So one hole must have two pigeons ,i am done .Is this correct ?

In general if we have $2n$ numbers and then we have $2n-1$ holes and $n+1$ pigeons ,But for $n=3$ onwards it is valid ? There is something wrong i think

Thanks

Suppose that you can select $n+1$ distinct integers from $\{1,\ldots,2n\}$ in such a way that all pairs differ by at least $2$. Let the chosen integers be denoted as $x_1,\ldots, x_{n+1}$. Also, let them be arranged in increasing order. Then, one has \begin{align*} x_1\geq&\,1,\\ x_2\geq&\,x_1+2\geq 3,\\ x_3\geq&\,x_2+2\geq 5,\\ \vdots&\\ x_k\geq&\,2k-1,\\ \vdots&\\ x_{n+1}\geq&\,2(n+1)-1=2n+1>2n. \end{align*} This is a contradiction, since $x_{n+1}$ is supposed to be at most $2n$.

Well, in your case when $n=5$, you shouldn't make the pairs $\lbrace1,2\rbrace,\lbrace2,3\rbrace,...,\lbrace8,9\rbrace,\lbrace9,10\rbrace$, but rather $\lbrace1,2\rbrace,\lbrace3,4\rbrace,...,\lbrace7,8\rbrace,\lbrace9,10\rbrace$, as in the first case the same integer appears in two different pairs.

When we have the correct pairs established, we can directly apply the pigeonhole principle on our $n=5$ "holes" $\lbrace1,2\rbrace,\lbrace3,4\rbrace,...,\lbrace7,8\rbrace,\lbrace9,10\rbrace$, and we see that if we have $n+1=6$ "pigeons" in the "holes", at least one of our "holes" must contain two pigeons.

We then apply the same reasoning to the general case.

You have nine holes and six pigeons, so the pigeonhole principle doesn't help because $6<9$.
Just use five holes $\{1,2\},\{3,4\},\{5,6\},\{7,8\},\{9,10\}$

Suppose to the contrary that there are no two consecutive numbers among the $n+1$ chosen numbers. Then there are $n$ "gaps" between the chosen numbers, with at least $1$ non-chosen number in each gap. But that would mean the total number of numbers is $\ge (n+1)+n$. This is impossible, since we only have $2n$ numbers.

Suppose not, then numbers differ by at least two. Order the numbers $a_1<a_2<\dots <a_n$. Then $a_1\geq 1$. We prove $a_i\geq 1+(i-1)2$ by induction

Base step $i=1$ is clear (we have to prove $a_1\geq 1+(1-1)2+1=1$)

Inductive step. We have $a_{i}\geq 1+(i-1)2$ and $a_{i+1}\geq a_i+2$ so $a_{i+1}\geq 1+(i-1)2+2=1+((i+1)-1)2$. Taking $i=n+1$ we get $a_n\geq 1+(n+1)2=2n+1$, a contradiction since all number must be in set $\{1,2,3\dots 2n\}$