Electrical – How to protect reed switches in dual float switch pump control circuit

This would be a very simple modification that might be acceptable.

^{simulate this circuit – Schematic created using CircuitLab}

When the switch is closed, \$Q_1\$ is off and the red LED is also off. When the switch is open, \$Q_1\$ receives base drive via the green LED and \$R_2\$ and \$R_3\$, sufficient to turn on the red LED. The green LED current will be cut down by a factor of about 20 (about \$1\:\textrm{mA}\$), so it won't be nearly so visible. There may also be some "leakage" into your pump/PIR circuit via \$R_3\$, but I don't think that will cause you trouble. You might want to lower \$R_1\$'s value, just in case. So this may, or may not, be acceptable in your circumstance. You'll have to decide. But it's easy to understand and implement.

Just for completeness, the "inverter" circuit that I believe was suggested might look like the following:

^{simulate this circuit}

But the price paid here is that the red LED current either goes through the red LED or else it goes through \$Q_1\$. But the current is always present. So you pay an added price in current draw. That may also be acceptable, though.