They still can't communicate.
The horizon you are talking about is the event horizon. Assuming a spacetime event happens here and now, and let light signal propagate forward in time. The wavefront of the light signal for $t = \infty$ or collapse time is the event horizon. For standard Friedman cosmology,
\begin{equation}
ds^2 = -dt^2 + R^2(t) \frac{dr^2}{1-kr^2} + r^2 d\Omega^2
\end{equation}
only $k = -1$ has event horizon.
Yet without algebra, spacetime diagram is enough to solve this problem.
- Bob should not have traveled towards Alice.
Suppose Bob sent a light signal 1 when he started. Then Bob has to travel slower than light, so he must fall behind signal 1. At any intermediate stage, he sent a light signal 2. Whatever the shape of light cones, signal 2 must arrive latter than signal 1. We can see this from the geodesic equation of the light ray $ds^2 =0 $, the slope
\begin{equation}
\frac{dt}{dr} = \frac{R(t)}{\sqrt{1-kr^2}}
\end{equation}
is increasing.
Alternatively, before signal 1 bypass signal 2, they must meet at the spacetime diagram. However the geodesic equation is general covariant, once they merge, they will be a single geodesic forever. Hence signal 1 will arrive no latter then signal 2. Actually, they can't merge at all, because the proper distance of these two light signals will keep fixed.
- Alice can't help Bob to speed up the signal.
Let's analyze signal 1. By assumption, Alice was within the event horizon of Bob. Hence (without any technical problem of course ) she would receive the signal. But the new signal she sent out had the same trajectory on spacetime diagram as if signal 1 were never been blocked. The reason is the trajectory of the light signal is determined by geodesic equation
\begin{equation}
ds^2 = 0 = g_{\mu\nu} dx^{\mu}dx^{\nu} = g_{\hat{\mu}\hat{\nu}} dx^{\hat{\mu}}dx^{\hat{\nu} }
\end{equation}
The proper distance is invariant under Lorentz transformation. And a bonus for light ray, the proper distance is zero, it is actually general covariant, which means the slope of the trajectory at any point is independent of the coordinate system we are using.
The light signal forwarded by Alice would propagate as if it was the original signal 1. By assumption, Charlie was outside the event horizon of Bob, the light signal will never reach him.
- Event horizon is shrinking due to expansion
We can see that the signal forwarded by Alice still couldn't reach Charlie, is it a contradiction?
No.
Charlie was in the event horizon of Alice at $t =0$, but at the time when Alice forwarded the massage, Charlie went outside of Alice's event horizon due to the expansion of universe. In other words, event horizon is shrinking. As I pointed above, the slope
\begin{equation}
\frac{dt}{dr} = \frac{R(t)}{\sqrt{1-kr^2}}
\end{equation}
is increasing.
I implicitly assumed the universe is expanding. If the universe is shrinking(as is possible for $k =-1$ ), i.e. $R(t) < 0$, then it's possible for Bob to communicate with Charlie. Because by in that case, the event horizon is expanding, and by traveling for a period effectively delay the emission of the signal. Charlie will eventually come into Bob's event horizon, then they can communicate with each other. But the method of ``forwarding messages'' is still of no help.
