Riding into a headwind faster than my top speed - how is it possible?
You can do it, because your bike is connected to the ground.

The work done when moving an object is proportional to distance and resistance force (which consists of air resistance and rolling resistance for bikes). The air resistance depends on air speed (ground speed + wind speed), but distance depends only on ground speed. Going slower reduces the energy spent, even when air speed remains high.

As an extreme example, consider just standing in place in wind. You are not doing any work at all, but your air speed is still high. When going against wind, you will soon find yourself doing more work. And when going downwind, you are doing negative work!

For objects not connected to ground things get different. A balloon in 40 MPH wind would drifting 40 Mph downwind in ground coordinates and would have to use the same power to stay in place in ground coordinates as it would use to move 40 Mph in still air.

A numerical example may help. We know the power equation for bicycles. For a bicycle on flat ground, at steady speed so that there is no acceleration or deceleration, in calm wind conditions, for typical rolling and aerodynamic coefficients of drag for a flat bar bike (Crr ~ .005 and CdA ~ 0.4 m^2), and a total mass of 85 kg, the power required to go 30 mph (13.33 m/s) is: 

.005 * 85 * 9.8 * 15.56 + 0.5 * 1.2 * 0.4 * 13.33 ^3 = 625 watts. 

However, what would the required power be to travel at 5 mph (2.22 m/s) into a 30 mph headwind? In that case, airspeed will be 13.33+2.22 = 15.56 m/s but ground speed will be only 2.22 m/s.

.005 * 85 * 9.8 * 2.22 + 0.5 * 1.2 * 0.4 * (15.56^2) * 2.22 = 140 watts.