Here is a counterexample, with the Heisenberg group. Define by induction $a_0=1$ and $a_{n+1}=(a_n^2+1)a_n$. Clearly $a_n$ tends to infinity.

Let $\Gamma$ be the Heisenberg group, consisting of matrices $m(x,y,z)=\begin{pmatrix}1&x&z\\ 0&1&y\\0&0&1\end{pmatrix}$ with $x,y,z\in\mathbf{Z}^3$. Let $C_n$ be the $n$-th congruence subgroup, namely those $m(x,y,z)$ with $x,y,z\in n\mathbf{Z}$. 

Let $\Gamma_n$ be the subgroup generated by the congruence subgroup $C_{a_n^2}$ and $m_n=m(1,a_n,0)$.

First check: $\Gamma_{n+1}\subset\Gamma_n$. Indeed $C_{a_{n+1}^2}\subset C_{a_n}\subset\Gamma_n$ because $a_n$ divides $a_{n+1}$. So it remains to check $m_{n+1}\in\Gamma_n$. Indeed, $m_n^{a_n^2+1}=m(a_n+1,a_n(a_n^2+1),0)=m(a_n,0,0)m_{n+1}$. Since $m(a_n,0,0)\in C_{a_n}$, we deduce $m_{n+1}\in\Gamma_n$.

Second check: $\bigcap_n\Gamma_n=\{m(0,0,0)\}$. Indeed, consider a nontrivial element $m(x,y,z)$. Choose $n$ such that $a_n$ is greater than $|x|,|y|,|z|$. Let us show that $m(x,y,z)\notin \Gamma_n$. Otherwise, it can be written as $m_n^km(aa_n^2,ba_n^2,ca_n^2)$, hence as $m(k+aa_n^2,ba_n^2,(ca_n+kba_n+k)a_n)$. Since the three coefficients are $<a_n$ in absolute value, we first deduce that $b=0$ so it equals $m(k+aa_n^2,0,(ca_n+k)a_n)$, then $k=-ca_n$, so it equals $m(a_n(aa_n-c),0,0)$, and finally it equals $m(0,0,0)$.

Finally, $m_n\in\Gamma_n$ is conjugate to the fixed element $m(1,0,0)$.