If the center of a finite group is trivial, are there two elements whose centralizers intersect trivially?

I believe that you can construct examples with arbitrary $k = k(G) > 1$ as follows.

Let $G$ be a semidirect product of an elementary abelian group $N$ of order $3^{2^k-1}$ with an elementary abelian $2$-group $H$ of order $2^k$, with action defined as follows.

There are $2^k-1$ subgroups of $H$ of order $2^{k-1}$, and each of the $2^k-1$ direct factors of $N$ is normalized by $H$, and centralized by one of these subgroups of $H$ of order $2^{k-1}$, where each factor has a different centralizer in $H$.

Then any subset of $G$ of order less than $k$ centralizes at least one of these direct factors of $N$.

To see that $k(G)=k$, let $H = \langle x_1,x_2,\ldots,x_k \rangle$, choose $y_1,y_2,\ldots,y_k \in N$ such that $[y_i,x_i] \ne 1$, but $[y_i,x_j]=1$ whenever $i \ne j$, and let $S = \langle x_1y_2,x_2y_3,\ldots x_ky_1 \rangle$. So the generators of $S$ all have order $6$, and since $x_1$ and $y_2$ are both powers of $x_1y_2$, etc, we have $H < S$ and also $y_i \in S$ for $1 \le i \le k$. Now $C_G(H) = H$ and $C_H(\langle y_1,y_2,\ldots,y_k \rangle) = \cap_{i=1}^k C_H(y_i) =1$, so we have $C_G(S)=1$.

Note that $G$ is a subgroup of the direct product of $3^{2^k-1}$ copies of the dihedral group of order $6$.

Here's another family of examples with arbitrary large $k$, more based on linear algebra. Fix any odd prime $p$. Consider the group $G_{p,s}$ of square matrices of size $s+2$ over the field $F=\mathbf{Z}/p\mathbf{Z}$ of the form $$m^\pm(u,v,z)=\begin{pmatrix}\pm 1 & ^tu & z\\ 0 & I_s & v\\ 0 & 0 & 1\end{pmatrix};$$ with $u,v\in F^s$ and $z\in F$. Its order is $2p^{2s+1}$. Its center is trivial (the center of the subgroup of index 2 is reduced to the cyclic group of elements $m^+(0,0,*)$.

For any $s-1$ elements in $G_{p,s}$. Then they are contained, for some hyperplane $H$ of $F^s$, in the subgroup $\Gamma$ consisting of those $m^\pm(u,v,z)$ with $u\in H$. Then there exists $w\in F^p\smallsetminus\{0\}$ such that $^tuw=0$ for all $u\in H$. Then $m^+(0,w,0)$ belongs to the centralizer of $\Gamma$. Hence $k(G_{p,s})\ge s$ (actually $\le s+1$).

The finite group $G=$ SmallGroup(486,176) in GAP or Magma notation has $k(G)=3$.