$\begin{align*}(\sqrt{n} + \sqrt{n+1})^2 &= 4n+2 - 2\left(n+1/2 - \sqrt{n(n+1)}\right) \\ &= 4n+2 - 2(AM(n,n+1) - GM(n,n+1)) \\ &\in (4n+1, 4n+2).\end{align*}$
(The first line is just algebra. In the second line, $AM$ and $GM$ are respectively the arithmetic and geometric means. To get the third line: $n < GM(n,n+1) < AM(n,n+1)=n+1/2$ by the AM-GM inequality, so $0 < AM(n,n+1)-GM(n,n+1) < 1/2$, and the third line follows.)
But there are no perfect squares between $4n+1$ and $4n+2$, and thus no integers between $\sqrt{n} + \sqrt{n+1}$ and $\sqrt{4n+2}$, QED.