Suppose $G$ is a finite group and there are $c$ conjugacy classes. I want to show that there is a non-identity element whose order is at most $c$.
I've tried using the class equation and seeing if there's a divisibility argument. But I'm not sure where to go from there. I also know conjugacy classes consist of elements whose orders are equal, so a contradiction argument might be possible. I would appreciate any hints.
Your idea of using the class equation is correct. Suppose the order of every element is larger than $c$. First try to prove that the center is trivial (hint: the center is an abelian subgroup). Then you would have $c-1$ non-trivial conjugacy classes. Call them $S_1,...,S_{c-1}$. Let $x_i$ be a representative from $S_i$. Then what would be the order of the centralizer $C_G(x_i)$? Try using the class equation to derive a contradiction from this.
(You should exclude the trivial group, of course...)
Since $|Z(G)|\leq c$ (each element of the center forms a singleton conjugacy class), if $Z(G)$ is nontrivial then you can take any nontrivial element $x\in Z(G)$; then $|x|\leq|Z(G)|\leq c$, and you are done.
So you may assume that the center of $G$ is trivial, in which case you have a singleton conjugacy class (the class of $e$), and $c-1$ conjugacy classes, each with more than $1$ element. If $S$ is one of these conjugacy classes, and $x\in S$, then we know that $|S|$ is the index of the centralizer of $x$; and we know that the centralizer of $x$ contains $x$; thus, $|S| = [G:C_G(x)]\leq [G:\langle x\rangle]$.
If every nontrivial element of $G$ has order greater than $c$, then letting $x_1,\ldots,x_{c-1}$ be representatives of the nontrivial conjugacy classes, we would have $$\begin{align*} |G| &= |Z(G)| + \sum_{i=1}^{c-1}[G:C_G(x_i)]\\ &\leq 1 + \sum_{i=1}^{c-1}[G:\langle x_i\rangle]\\ &= 1 + \sum_{i=1}^{c-1}\frac{|G|}{|x_i|}\\ &\lt 1 + \sum_{i=1}^{c-1}\frac{|G|}{c}\\ &= 1 +\frac{(c-1)|G|}{c}. \end{align*} $$ This means that $|G| \lt 1 + |G| - \frac{|G|}{c}$, so $\frac{|G|}{c}\lt 1$. This forces $G$ to have more conjugacy classes than elements... which is something of a problem. The contradiction arises from the assumption that every nontrivial element of $G$ has order greater than $c$, so we conclude that there exists $x\in G$, $x\neq e$ such that $|x|\leq c$, as desired.
