You practically have the solution. Let us arrange the selected integers in non-decreasing order and denote them $a_1, a_2, \ldots, a_{21}$.
Now, $0 < a_1 \leq a_{21} \leq 400$, therefore $\sqrt{a_{21}}- \sqrt{a_1} < 20$.
Now notice that
$$
\sqrt{a_{21}}- \sqrt{a_1} = (\sqrt{a_{21}} - \sqrt{a_{20}}) + (\sqrt{a_{20}} - \sqrt{a_{19}}) + \ldots + (\sqrt{a_2} - \sqrt{a_1}).
$$
There are $20$ summands on the right hand side, they are all nonnegative, and their sum is less than $20$. It follows by the pigeonhole principle that one of them is less than $1$.
This is somewhat different from the usual pigeonhole method, i.e. there aren't really any pigeons and pigeonholes. But the principle is similar. We assume that each summand is greater than or equal to $1$, we conclude that the sum is greater than or equal to $20$, and this is a contradiction.
UPDATE: if you look at BFD's answer, it becomes clear that he standard, usual, "normal" pigeonhole principle can be applied too.