Here is a short proof. Thanks to David Speyer for simplifying an earlier proof of mine (see the comments below).

For each colour $i \in [n]$, let $a_i$ be the number of vertices incident to an edge of colour $i$.  Observe that $a_i \geq 2\sqrt{n}$ for all $i$, where equality is obtained if and only if the edges of colour $i$ induce a $K_{\sqrt{n}, \sqrt{n}}$.  Thus, $\sum_{i \in [n]} a_i \geq 2n \sqrt{n}$.  On the other hand, $\sum_{i \in [n]} a_i$ is the number of ordered pairs $(v,i)$, where $v$ is a vertex, and $i$ is a colour incident to $v$.  Therefore, since there are only $2n$ vertices, some vertex must be incident to at least $\sqrt{n}$ colours.