Why is oxygen oxidized with lead instead of being reduced?
Your confusion stems from that fact that you are trying to look at what reactions are favorable instead of the reaction as written.

Oxygen is readily reduced because it is a strong oxidant. You can check any reduction half reaction with oxygen on the left side. Chances are that the reduction potential will be fairly positive.

In the reaction you have written, oxide is being oxidized to oxygen. Nowhere (including your textbook) do we claim that this reaction is favorable or spontaneous in the forward direction. In fact, if you consult a table of free energies of formation (example), you would find an entry like:

$$\ce{PbO (s, red)}, \Delta G^{\varnothing}_{f}=-189.3\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$
or
$$\ce{PbO (s, yellow)}, \Delta G^{\varnothing}_{f}=-188.5\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$

These values indicate that indeed, the reverse reaction is the one that's favorable, the one where oxygen serves as an oxidizing agent and is reduced.

However, that is not the reaction as written. The notation of chemistry allows us to describe reactions that are hypothetical or unfavorable.