Since $I$ and $J$ are constrained to be nonempty, the assertion may in fact be false if one of the $A_i$'s is empty (for example, $n=2$, $A_1=\{1\}$, $A_2=\{2\}$, $A_3=\emptyset$.) On the other hand, if at least two $A_i$'s are empty (say $A_i$ and $A_j$), we win by choosing $I=\{i\}$ and $J=\{j\}$. So we will assume henceforth that all of the $A_i$'s are nonempty, and we prove the result under this assumption.
Write down the "incidence matrix" $M$, which has $n+1$ rows, corresponding to the $A_i$'s, and $n$ columns, corresponding to the elements of the set $A$, as follows: in the row corresponding to the set $A_i$, put a $1$ in each column corresponding to an element of $A_i$, and zeros in the remaining columns. As there are more rows than columns, the rows must be linearly dependent, so there's a row vector ${\vec c}=(c_1,\ldots,c_{n+1})$ such that ${\vec c}\, M=0$, with the $c_i$'s not all zero.
Define $I=\{ i\mid c_i>0\}$ and $J=\{j\mid c_j<0\}$. If ${\vec a_i}$ denotes the row vector corresponding to the set $A_i$, we then have $$\sum_{i\in I}c_i {\vec a_i}=\sum_{j\in J}|c_j| {\vec a_j}.$$ Call this sum $\vec u$, and note that all coefficients in the sum are positive. Since none of the $A_i$'s are empty, $\vec u$ is a nonzero vector with nonnegative coordinates, so both $I$ and $J$ are nonempty (and they're clearly disjoint.)
A coordinate in the left-hand sum is nonzero precisely when it is nonzero in some $\vec a_i$ with $i\in I$, so the nonzero coordinates of $\vec u$ correspond precisely to the elements of $\cup_{i\in I}A_i$. But the same is true on the right-hand side, so we conclude that $\cup_{i\in I}A_i = \cup_{j\in J}A_j.$
