Understanding the voltage across the inductor in a boost converter
What makes the voltage across the inductor stay constant at Vi?

It stays constant because the incoming supply defines that voltage but, if you left the inductor connected across the supply, after a few tens of microseconds, the current might be tens or hundreds of amps and that will cause the incoming supply to collapse. Then you're in trouble.
But that isn't how a boost circuit works. The inductor is placed in parallel with the supply and current ramps up to &quot;several&quot; amps. That peak current is, in effect, representing the stored energy in the inductor. Microseconds later, the inductor is disconnected from the incoming supply and, that stored energy is released into the output capacitor and load. Then the process starts again.
The point is that you don't hold the inductor connected to the incoming supply for more than a few microseconds (or a few tens of microseconds in really powerful boost converters). Here's an example of a boost converter's inductor current: -

Web boost-regulator calculator.
On the example above, current ramps up from 0 amps to 15.656 amps in 6.132 μs. It then transfers the energy (during which it ramps down) in 2.044 μs and, a little while later (1.824 μs), the cycle begins again. This is just one example and, we'd expect the incoming supply to be able to deliver 15.656 amps without drooping too much. This boost converter is operating in discontinuous conduction mode (\$\color{red}{\boxed{\text{DCM}}}\$) because the inductor current falls to zero before the cycle restarts.
A larger value inductor (10 μH from 4.7 μH) means the current peak is less: -

Peak current is 10.9 amps and now the switching circuit is operating in what is known as continuous conduction mode (\$\color{blue}{\boxed{\text{CCM}}}\$) where the inductor holds on to a little bit of energy as the cycle starts again