... $\sum c_{n_i}$ is absolutely convergent.
The best I can show is $\sum c_{n_i}$ converges. This is because $\sum c_n$ being absolutely convergent implies it is convergent. If we define a sequence of partial sums of $c_n$, we can construct a subsequence of this sequence that corresponds to $\sum c_{n_i}$, thereby showing the latter expression converges. But, this does not imply absolute convergence.
On
Since $\sum_{n=1}^{\infty} c_{n}$ is absolutely convergent then the sequence $(\sum_{k=1}^{n} |c_{k}|)_{n=1}^{\infty}$ of partial sums is convergent. This is equivalent to the convergence of the tail sequence $(\sum_{k=n}^{\infty} |c_{k}|)_{n=1}^{\infty}$ to zero. That is, given any positive number $\epsilon$, there exits $N_{\epsilon}$ such that $$|\sum_{k=n}^{\infty} |c_{k}| |<\epsilon$$ for all $n >N_{\epsilon}.$
So, if $(c_{n_{k}})_{k\geq 1}$ is a subsequence of $(c_{n})_{n\geq 1}$ then
$$|\sum_{k=n}^{\infty} |c_{n_{k}}| |<\epsilon$$ for all $n >N_{\epsilon}$ because, by the definition of a subsequence, $n_{n}\geq n$ for all $n\geq 1$.
Hence the sequnce $(\sum_{k=n}^{\infty} |c_{n_{k}}|)_{n=1}^{\infty}$ converges to 0, and equivalently $(\sum_{k=1}^{n} |c_{n_{k}}|)_{n=1}^{\infty}$ is convergent.
Let $C = \sum_{n=1}^\infty |c_n| < \infty$. Then $\sum_{i=1}^k |c_{n_i}| \leq C$ for any $k$ so the partial sum sequence for $\sum |c_{n_i}|$ is monotone increasing and bounded above.
However, unless I'm mistaken, your proof that $\sum c_{n_i}$ converges doesn't quite work. The problem is that you can't extract the partial sum sequences of $c_{n_i}$ as a subsequence of the partial sum sequence for $c_n$ in general. To see this, note that the partial sum sequences for $c_{n_i}$ could, for example, look like $$c_2 + c_4 + c_6 + \dots + c_{2k}$$ whilst any subsequence of the partial sum sequence for $c_n$ will involve terms of the form $c_{2n+1}$ (and in particular $c_1$).
Of course, convergence of $\sum c_{n_i}$ does follow from the absolute convergence shown above.