Prove that three $2\times2$ matrices that commute are linearly dependent

Commuting matrices are simultaneously triangularisable. We may assume $A$, $B$ and $C$ are upper triangular. If they are linearly independent, they span the three-dimensional space $T$ of upper triangular matrices. Therefore all elements of $T$ commute. But they don't.

It is hard to beat Lord Shark's answer, but we may prove the statement without knowing that $A,B,C$ can be simultaneously triangularised.

Proof. If they are linearly independent, we can find two linearly independent traceless matrices $X$ and $Y$ in their linear span. Hence we may assume that $X$ is either $\operatorname{diag}(-x,x)\ (x\ne0)$ or a nilpotent Jordan block. Any traceless matrix that commutes with $X$ is thus a scalar multiple of $X$. This contradicts the property of $Y$.