Why is hydrated beryllium ion acidic?
What is meant by a &quot;very strong&quot; Be−O bond? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size?

You got that absolutely right.
$\ce{Be\bond{-}O}$ is a strong bond because of the small size of Be. Smaller cation size means a stronger pull on the $\ce{O}$ electrons, thus reducing the bond length and hence increasing the bond strength.

[...]the electrons in the O−H bonds would get shifted towards oxygen. But as far as I can tell this only makes the $\ce{O\bond{−}H}$ bonds more polar.

You struck at the right point. Let's look at the original question.
I'm sure you must know that a stronger acid has a higher tendency to release $\ce{H+}$ ions. This means that the electron pair should be drawn away from $\ce{H}$ atom in the $\ce{O\bond{−}H}$ bond. Thus, a polarized bond makes it easier for the H+ ion to 'escape' from the molecule.

Why does a polar bond have to be a weaker one?

It is not appropriate to answer this question in definitive terms. The strength of a bond is defined by its bond length, not its polarity. To address the confusion you're having, look back at the context of the question. $$\ce{[Be(H2O)4]^{2+} + H2O -&gt; [Be(H2O)3OH]+ + H3O+}$$
We're talking about a compound with polar O-H bonds dissolved in water. Water has a high dielectric constant, and can easily break ionic/polar bonds since it liberates a high amount of solvation energy. This is because of its high dielectric constant.
So, are polar bonds weaker than non-polar bonds in water? Sure, I can see your line of thinking. But, always remember that ionic bonds can easily be broken by water, but that doesn't mean that ionic bonds are weaker than non-polar bonds.