Very cute indeed! Suppose that $x_1 + x_2 + \cdots + x_{a_{i+1}} = a_i + \epsilon_i$ for all $i \geqslant 1$. Assume, by induction that the statement is true for all $(i,j)$ with smaller sum. Note that by induction, $a_j \leqslant a_{i+j-1} - a_{i-1} = (x_1 + x_2 + \cdots + x_{i+j} - \epsilon_{i+j}) - (x_1 + x_2 + \cdots + x_{i} - \epsilon_i) = (x_{a_i+1} + x_{i+2} + \cdots + x_{a_{i+j}}) + (\epsilon_i - \epsilon_{i+j}) < a_{i+j} - a_i + 1$. Since the $a_k$ are integers, this forces that $a_{i+j} \geqslant a_i + a_j$, as desired.