I found the solution on my own, so I share it with you.
We know that
\begin{equation}
\gamma_X(h) = \sigma^2\sum_{i = 0}^\infty c_ic_{i+h}.
\end{equation}
To test the convergence consider
\begin{equation}
\frac{1}{\sigma^2}|\gamma_X(h)| \leq \sum_{i=0}^\infty|c_i|\cdot|c_{i+h}| \leq \sqrt{\sum_{i=0}^\infty|c_i|^2}\cdot\sqrt{\sum_{i=0}^\infty|c_{i+h}|^2} = ||c_i||_{\mathcal{l}^2}\cdot||c_{i+h}||_{\mathcal{l}^2}.
\end{equation}
As $(c_i)$ is a member of the $\mathcal{l}^2$ space, so $||c_i||<\infty$. What remains is to check $\lim_{h\rightarrow\infty}||c_{i+h}||_{\mathcal{l}^2}$.
\begin{equation}
||c_{i+h}||_{\mathcal{l}^2}^2 = \sum_{i=0}^\infty|c_{i+h}|^2 = \sum_{i=h}^\infty|c_i|^2 \rightarrow 0,\;\;\text{as}\;\;h\rightarrow\infty.
\end{equation}
Which holds because the $\mathcal{l}^2$ space is complete and tails of convergent series converge to zero.
Hence
\begin{equation}
|\gamma_X(h)| \leq \sigma^2 ||c_i||_{\mathcal{l}^2}\cdot||c_{i+h}||_{\mathcal{l}^2} \rightarrow 0,\;\;\text{as}\;\;h\rightarrow\infty.
\end{equation}