Diode not behaving as short circuit
Initially, the inductor resists the change of current, making the diode the path of least resistance and causing it to carry most of the current. When the magnetic field in the inductor builds up, the voltage across it decreases as it allows more current to pass. The diode has a forward voltage drop (typically 0.6V) to account for, so it won't conduct any current after the voltage across the inductor falls below the diode's forward voltage.
Yes the previous posters are right. To further clarify, a diode is not a short circuit but a threshold device, it starts conducting whenever voltage across it (when oriented properly to conduct) is greater than some value, typically 0.6V (but may differ for special types).
So it behaves like this whenever the voltage is lower than 0.6 V no current will flow and when the voltage is over this threshold current flows.
The inductor responds to sudden changes in current in a different way, it exhibits something called impedance, that is a way to say that while it has a resistance R it also has an inductance L, a component that is directly dependent on frequency.
So an inductor when suddenly connected or disconnected from a voltage supply reacts by spiking up voltage for a brief while and the current is initially almost zero, only to settle a brief moment later with smaller currents and voltages approaching zero.
The diode in the circuit sees this increase in voltage (while the current is still almost zero in the coil) and it closes, letting the spike flow through it, reducing also the excessive voltage on the coil and thus the large current in the diode that flows for a very short time.
A very common arrangement usually called a SNUBBER is what you will find in some switching relays or even solid state devices. Its function is to stop the excessive voltage spike from breaking coil insulation by conducting temporarily the large voltage spike and then to close as the voltage on the coil returns close to zero. I merely translated the above equations and observations in layman terms, hope it helps.
When enough time has passed, the inductor behaves as a short circuit and bypasses the diode. That means \$V_d=0\$ and no current flowing through the diode.
OTOH, at the precise time of switching, the inductor will try to keep its current (that happens to be zero because it is not energized). Because of that, it briefly behaves like an open circuit and all current flows through the diode, which will be forward biased until the voltage in the inductor falls below \$V_f\$.