How to write the redox reaction of chlorate to chlorine and oxygen using half rections?
This is tricky and one way to think about it that I can offer is to treat $\ce{ClO3-}$ as a combination of $\ce{Cl^5+}$ and $\ce{O^2-}$. Since the problem is that the starting material is undergoing disproportionation with different elements being reduced/oxidised, if you could split the starting material into the two different elements, then you could circumvent the issue. Obviously the bonding in the chlorate ion is not ionic but this procedure allows you to construct "half-equations" and then an overall equation.

I would not suggest that you write this in an exam.

The two half-reactions are then:

$$\begin{align}
\ce{Cl^5+ + 5e-} &amp;\longrightarrow \ce{1/2Cl2} \tag{1} \\
\ce{O^2-} &amp;\longrightarrow \ce{1/2 O2 + 2e-} \tag{2}
\end{align}$$

$[(1) \times 2] + [(2) \times 5]$ gives

$$\ce{2 Cl^5+ + 5O^2- -&gt; Cl2 + 5/2 O2}$$

In order to "re-form" your chlorate, you need three oxide ions per Cl ion, so you need to add one more oxide to both sides:

$$\ce{2 Cl^5+ + 6O^2- -&gt; Cl2 + 5/2 O2 + O^2-}$$

Now you can get back your chlorate:

$$\ce{2 ClO3- -&gt; Cl2 + 5/2 O2 + O^2-}$$

Now it's highly unlikely that there are going to be free oxide ions hanging around, so we just protonate them by adding $\ce{2 H+}$ on both sides. The oxide ion simply becomes water:

$$\ce{2 ClO3- + 2H+ -&gt; Cl2 + 5/2 O2 + H2O}$$


  I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine (V)'s so you have to use a water to make the charge balance work out.


Yeah, if all six oxygens became $\ce{O2}$ then they would lose 12 electrons. Therefore, one of the oxygens does not become $\ce{O2}$; it becomes $\ce{H2O}$ and remains as oxygen(-2). It is not about charge balance.