We will prove this by contradiction. Suppose that in the interval $[A,B]$, there are no perfect squares, e.i. there exists a positive integer $m$ such that $m^2 \leq A < B \leq (m+1)^2$. Notice that in the interval $[A,B]$, there are $B - A + 1 = x_n + 1 = \frac{n(n+1)}{2} + 1 = \frac{n^2 + n + 2}{2}$ numbers. But in the interval $[m^2,(m+1)^2]$, there are $(m+1)^2 - m^2 + 1 =2(m+1)$ numbers, so we must have that $\frac{n^2 + n + 2}{2} \leq 2(m+1)$ $\implies n^2 + n \leq 4m + 2$ $\implies n^2 + n - (4m+2) \leq 0$ $\implies \Delta = 1^2 + 4(4m+2) \leq 0$ $\implies m <0$ $\implies B < A$; contradiction. Therefore, there exists at least one square in the interval $[A,B]$. QED