Kk I think the following works. Let's call our 7 cards $c_1, \ldots c_7$. View the "symbols on the card $c_j$" as an element $v_j$ of $\mathbb{F}_2^{\oplus 6}$: namely, if our 6 possible symbols are $s_1, \ldots s_6$, let the $i^{th}$ coordinate $v_j$ be the number of times $s_i$ appears on the card.
For example, if $c_1$ reads: $s_1 s_3 s_5$, $v_1 = (1, 0, 1, 0, 1, 0)$.
Now take your 7 cards, and look at their corresponding vectors side by side, i.e. form the matrix $$\begin{pmatrix} v_1 & v_2 & \ldots & v_6 & v_7 \end{pmatrix}$$This gives us a map from $\mathbb{F}_2^{\oplus 7} \rightarrow \mathbb{F}_2^{\oplus 6}$: it must have a nonzero kernel.
Suppose $w \neq 0$ is in the kernel-by construction the subset of the 7 cards corresponding to in which coordinates $w$ has a 1 is your desired subset (i.e. each symbol occurs an even number of times).
Lemme know if you have qs :p (cool question btw, thanks for sharing!)