Pigeonhole principle: prove that any set of $n+1$ integers from $\{1,2,...,2n\}$ has two consecutive integers differ by one.

Let $n \ge 1$ be an integer. Use the Pigeonhole Principle to prove that in any set of $n + 1$ integers from $\{1, 2, . . . , 2n\}$, there are two integers differ by one.
I've attempt and come up with this: Let S be $\{1,2,...,2n\}$, $T_0$ be a subset of S with $n$ elements, $T_1$ is a subset of S with $n+1$ elements. We form $T_1$ by adding a number to $T_0$. If $T_0$ already has two integers differ by one, then we're done. If $T_0$ has no integers differ by one, how do we know $T_1$, having 1 more number, has two integers differ by one?

Hint:

There are $n$ holes in $\{1, 2, \cdots, 2n\}$, namely, $\{1, 2\}$, $\{3, 4\}$, $\cdots$, $\{2n - 1, 2n\}$.