Why do double bonds break
Keep in mind, that the reason reactions occur spontaneously is because there's an overall driving force: $\Delta G = \Delta H - T\Delta S$. Since there are fewer molecules on the right hand, we can guess that $\Delta S$ isn't driving this reaction. ($G$ is Gibbs free energy, $S$ is entropy, $H$ is enthalpy, and $T$ is temperature.)

So then your question boils down to "why is this reaction favorable for $\Delta H$?"

Even before I look up the bond enthalpies for each of the bonds involved, I have an intuition. Yes, we're breaking the double bond in cyclohexene. We're also breaking the $\ce{Br-O}$ bond in $\ce{Br-O-H}$. So that's two bonds that are broken. 

You're right -- that costs energy, so there's clearly something more energetically favorable on the other side. Well, in the product, I'm forming a $\ce{C-Br}$ and a $\ce{C-OH}$ bond.

So the net reaction breaks two bonds, and forms two bonds. As you say, the general intuition is that the "double" part of a double bond isn't quite as strong as the standard $\sigma$ single bond.

Let's look up the bond enthalpies also through Wikipedia:


$\ce{C=C}$ (vs. $\ce{C-C}$ which remains in the product): ~146 kcal/mol vs. ~84 kcal/mol = ~62 kcal/mol
$\ce{Br-O}$ = ~56 kcal/mol (I had to look this one up, so it's a bit iffy)
$\ce{C-Br}$ = ~69 kcal/mol
$\ce{C-O}$ = ~85 kcal/mol


Net $\Delta H \approx$ -36 kcal/mol.

Even if some of the bond enthalpies are off a bit, it's clearly energetically favorable. It has nothing to do with how the species are mixed, or really the temperature. (Well, my guess is that $\Delta S$ is not favorable here, since two molecules form one, so high temperature might make the reaction less favorable from a $\Delta G$ standpoint.) Nor do you need UV light or anything like that -- the difference in bond strengths will push this reaction.