Let $G$ be a finite $p$-group. It's an old result of Suzuki that if $G$ possesses a self-centralizing subgroup of order $p^2$ then $G$ has maximal class, and in fact the converse is true. However, the proof of the converse appears not particularly easy, at least as far as I know. See Huppert, Section III.14, or Leedham-Green--McKay for definitions. I will refer to Huppert for specific results.
The only proof I know is the following: if $x\in G$ lies outside all $2$-step centralizers $K_i=C_G(\gamma_i(G)/\gamma_{i+2}(G))$ for $2\leq i\leq n-2$ then $C_G(x)$ has order $p^2$ by Huppert, III.14.13. This proof is easy, just an induction up the central series. Thus the problem is proving that not all maximal subgroups are $2$-step centralizers. If $|G|$ is small, say around $|G|\leq p^{p+1}$, then there are not enough $K_i$ for all maximal subgroups to be one of the $K_i$. If $|G|$ is large though, one appears to have to use the fact that $G$ cannot be exceptional, i.e., all $K_i$ are in fact equal (Huppert, III.14.18), or at least $G/Z(G)$ is not exceptional and hence the $K_i$ are at most two maximal subgroups (Huppert, III.14.6).
So my question is, if one is only trying to prove the existence of self-centralizing subgroups of $p$-groups of maximal class, is there a faster way that doesn't involve using the lack of exceptionality?
One naive approach is to use induction on $|G|$. The result holds for $G/Z(G)$, so you obtain a self-centralizing subgroup there. The preimage is a subgroup $B$ of order $p^3$, containing $Z(G)$. If this subgroup is non-abelian then any subgroup $A$ of order $p^2$ other than $Z_2(G)\leq B$ is self-centralizing. But if $B$ is abelian, this doesn't work. And I don't see why $B$ is not abelian. Equivalently, $B$ can be contained in the $2$-step stabilizer $C_G(Z_2(G))$, and I don't see how to force it not to be, or at least that there is a choice of it not being. It seems that I end up back at trying to make sure there are enough maximal subgroups that are not the $K_i$ to allow me to make a choice, but I cannot see how to do this without proving the much stronger results about exceptionality.