Why do buck converters end up in steady state?
Why doesn't the current in the inductor gradually increase in each cycle until it reaches infinity?

Because your circuit has a load that is not a short circuit.
\$ di/dt = e/L \$
When the FET is ON, voltage across the inductor is Vin-Vout, so its current increases according to \$ di/dt = (V_{in}-V_{out})/L \$.
When the FET is OFF, voltage across the inductor is -Vout, so its current decreases according to \$ di/dt = (V_F-V_{out})/L \$, with VF from the diode.
In both cases, I neglected resistance in the circuit, but you'd have to account for it if you want more accurate results.
If there is a load that isn't a short circuit on the output, like your resistor, then output voltage will rise. As Vout rises, during on-time, inductor current increases less (because that part is proportional to Vin-Vout), and during off-time it decreases more (because that part is proportional to Vf-Vout).
At some value of Vout, the increase in current during ON-time is equal and opposite to the decrease in current during OFF-time, so the inductor current is a sawtooth: that's steady state.

If there is an output cap and no PWM soft-start there will be some LC ringing and inductor current can get quite large as the output cap charges:

Since you use a non-synchronous buck if the inductor current never drops below zero it will be in continuous conduction mode, but if it drops below zero the diode will turn off at this point, so it will be in discontinuous mode:

If the output is shorted, Vout remains 0V, so inductor current will increase each cycle, but during the OFF-time portion of the cycle, it won't decrease much, since there is only the diode Vf across it. So it will keep rising until the FET or the diode overheat. I've run the simulation with a 500mA diode, so it would smoke rather fast at 50 Amps:

If there is no load on the output, just a cap, then output voltage will rise to be almost equal to the input voltage, and when the FET turns on, inductor current will not increase since (Vin-Vout)=0. In this sim, even though the PWM is 50%, Vout goes very close to Vin, because once it's there, there is no way for the cap to discharge without a load: