This is actually not true. For a counterexample, take $(a_1,a_2,a_3) = (1,\phantom. \pi,\phantom.\pi/(\pi+1))$ and $(c_1,c_2,c_3) = (0,0,1)$, so our sequences are
$\lbrace n_1 \rbrace$,
$\lbrace n_2 \pi \rbrace$, and
$\lbrace 1 + n_3 \frac\pi{\pi+1}\rbrace$.
If $n_1$ is near $n_2 \pi$ then it's also near $(n_1+n_2)\frac\pi{\pi+1}$, and therefore far from $1 + n_3 \frac\pi{\pi+1}$ for any integer $n_3$.

As this example suggests, the correct condition (for any number of arithmetic progressions) is that the *reciprocals* of the common differences $a_i$ be linearly independent over ${\bf Q}$.

Indeed, given nonzero $a_1,\ldots,a_{k+1} \in {\bf R}$, consider the image of the lattice $L = \oplus_{i=1}^{k+1} {\bf Z} a_i \subset {\bf R}^{k+1}$ under the quotient map $\pi : {\bf R}^{k+1} \rightarrow V := {\bf R}^{k+1} / {\bf R}v$ where $v=(1,1,\ldots,1)$. Then the desired result is true for all $c_1,\ldots,c_{k+1}$ **iff** $\pi(L)$ is dense in $V$. But $\pi(L)$ is a subgroup of $V$, so its closure is a closed subgroup — and a closed subgroup $G \subseteq V\phantom.$ is properly contained in $V\phantom.$ **iff** there a nonzero linear functional $\varpi: V \rightarrow {\bf R}$ such that $\varpi(G) \subseteq \bf Z$. Hence $\pi(L)$ fails to be dense in $V\phantom.$ **iff** there is a nonzero functional $V \rightarrow {\bf R}$ taking $\pi(L)$ to a subset of ${\bf Z}$. But the linear functionals on $V\phantom.$ are simply the functionals on ${\bf R}^{k+1}$ vanishing on $v$; that is, the maps $(x_1,\ldots,x_{k+1}) \mapsto \sum_{i=1}^{k+1} b_i x_i$ for some $b_i\in\bf R$, not all zero, such that $\sum_{i=1}^{k+1} b_i = 0$. Hence our condition is $b_i a_i \in \bf Z$, or equivalently $b_i \in {\bf Z} a_i^{-1}$. There exist such $b_i$ that sum to zero **iff** the $a_i^{-1}$ are linearly dependent over the rationals, **QED**.