[evan_au]

Closing the switch should mean no current flows, thanks to the infinite time constant of the superconductor

A characteristic of Superconductivity is that it has zero DC resistance.
 
If your circuit has a single turn, it will have minimal inductance, and current will rise extremely quickly (assuming that your power source has zero impedance). This is not pure DC.

Superconductors do not work so well with rapidly-changing magnetic fields, as flux vortices dissipate some of the energy.
- That's why the various superconductive power distribution systems tested around the world tend to deliver DC rather than AC.

https://encyclopedia.pub/entry/9252

[evan_au]

Closing the switch should mean no current flows, thanks to the infinite time constant of the superconductor.

If we assume the inductance is around 100nH (for a circuit with length 4 inches), and the inductance is the main impediment to current flow...

The current can be described as V=L ^dI/_dt
- Where V is the applied voltage. We'll use 1V DC in this example
- L is the inductance in Henries (but we will use 100nH in this example = 10^-7 H)
- I is the current flowing through the circuit
- ^dI/_dt is the rate of increase of current with time, in Amps per Second.

So V=L ^dI/_dt
1 = 10^{-7 dI}/_dt
or ^dI/_dt = 10⁷ Amps/second = 10 million Amps after 1 second, which is a huge amount of current!
This is enough to turn most thin superconductors into normal conductors, due to exceeding the critical current.

So the assumption that "no current flows" is a fallacy.