Represent each of the seven cards by a binary vector: the vector for card $i$ is $v_i=\langle b_{i1},\dots,b_{i6}\rangle$, where $b_{ij}=1$ if the $j$-th symbol appears on card $i$, and $b_{ij}=0$ otherwise. For $j=1,\dots,6$ and $\varnothing\ne S\subseteq\{1,\dots,7\}$ let $b^S_j=\left(\sum_{i\in S}b_{ij}\right)\bmod 2$, and let $v_S=\left\langle b^S_1,\dots,b^S_6\right\rangle$; the problem is to show that $v_S=\vec0$ for some non-empty $S\subseteq\{1,\dots,7\}$.
For each non-empty $S\subseteq\{1,\dots,7\}$ let $C(S)=\big\{j\in\{1,\dots,6\}:b^S_j=1\big\}$. There are $2^6=64$ possible values for $C(S)$. On the other hand, there are $2^7-1=127>64$ non-empty subsets of $\{1,\dots,7\}$. Thus, there are distinct non-empty $S,T\subseteq\{1,\dots,7\}$ such that $C(S)=C(T)$. It should be clear that if $S\cap T=\varnothing$, then $C(S\cup T)=\varnothing$, and therefore $v_{S\cup T}=\vec0$. What if $S\cap T\ne\varnothing$?
Fix $j\in\{1,\dots,6\}$. Then, with all arithmetic done mod $2$, we have
$$\begin{align*}
0&=1+1\\
&=\sum_{i\in S}b^S_i+\sum_{i\in T}b^S_i\\
&=\sum_{i\in S\setminus T}b^S_i+\sum_{i\in T\setminus S}b^T_i+\sum_{i\in S\cap T}\left(b^S_i+b^T_i\right)\\
&=\sum_{i\in S\setminus T}b^S_i+\sum_{i\in T\setminus S}b^T_i+\sum_{i\in S\cap T}\left(1+1\right)\\
&=\sum_{i\in S\setminus T}b^S_i+\sum_{i\in T\setminus S}b^T_i\\
&=\sum_{i\in S\Delta T}b^{S\Delta T}_i\;,
\end{align*}$$
where $S\Delta T$ is the symmetric difference of $S$ and $T$. But then $C(S\Delta T)=\varnothing$, so $v_{S\Delta T}=\vec0$. (Note that $S\Delta T\ne\varnothing$, since $S\ne T$.) Of course $S\Delta T=S\cup T$ when $S\cap T=\varnothing$, so we didn’t really need the nice special case except as a pointer towards the more general result.
Note that a pigeonhole argument also underlies uncookedfalcon’s nice linear algebraic solution.
