Your charge distribution is correct, though I usually don't focus on using a "correct" $\pm$ distribution in capacitors--if your sign is incorrect, you get a negative value of charge on the + plate. No biggie. In complex situations, it is sometimes even impossible to predict cjarge distribution without solving the circuit.
And your book is incorrect. $C_1-C_2$ are in series, though the entire branch is in parallel with $C_3$ and its opposite branch.
Parallel is when the current is split, while series is when current is constant. Series is a single wire with an in and an out, with components along the wire. Parallel is when there are many wires with their ends twisted together. Current goes in/out through the twisted ends.
You seem to have grasped that, though :)
Out of curiosity, which book is this? (the capacitors look like they're from Resnick)
Suppose you imagine the battery to be a variable voltage, and start with the voltage at zero. Obviously everything is uncharged.
Now turn the battery up to 1V. As you do this positive charge leaves the positive terminal and an equal and opposite negative charge leaves the negative terminal. We know the charges leaving the positive and negative terminals must be the same because the battery is a conductor and can't develop a net charge like a capacitor. Let's call the charge that leaves the battery $Q$.
The only place the charge that leaves the battery can go is onto the capacitors, so both capacitors now have a charge of $Q$ on them. We know that for a capacitor of capacitance $C$, the voltage across the capacitor is given by:
$$ V = \frac{Q}{C} $$
Call the voltage of the top (1$\mu$F) capacitor $V_1$, and the voltage of the bottom (2$\mu$F) capacitor $V_2$. Then:
$$ V_1 = \frac{Q}{C_1} $$
$$ V_2 = \frac{Q}{C_2} $$
Dividing the first equation by the second plus a bit of quick rearrangement gives:
$$ V_1 = \frac{C_2}{C_1} V_2 $$
The two voltages must add up to 1V because we have a 1V battery, therefore:
$$ V_1 + V_2 = 1$$
If you substitute for $V_1$ you get:
$$ \frac{C_2}{C_1} V_2 + V_2 = 1$$
and dividing through by $(1 + C_2/C_1)$ gives:
$$ V_2 = \frac{1}{1 + C_2/C_1} $$
Tidy this up by multiplying to top and bottom of the right hand side by $C_1$ and you get the equation you're trying to prove:
$$ V_2 = \frac{C_1}{C_1 + C_2} $$
Just to check, feed in $C_1 = 1$ and $C_2 = 2$ and $V_2$ does indeed come out as 1/3V.
Best Answer
Here's what I do (or rather, what I just did) to convince myself of your result.
We're trying to "reduce" the two capacitors in series to an equivalent capacitor. Equivalent in every way, including how much charge would flow when discharging the capacitor.
Imagine discharging the two capacitors that are series. How much total charge would flow? Since, as you've stated, the "input" and "output" sides store equal magnitude charges $|Q_1|=|Q_2|\equiv Q$, then we should be able to recover this much charge during discharging.
By discharging the equivalent capacitor, then, one should expect to have a total charge $Q$ flow during discharging. Thus, $Q_\mathrm{equiv}=Q$.