I misunderstood; you're on board with the technique albeit a 5-diode approach. Anyway, since the application is anti-chatter or protecting against dropouts measured in the few milliseconds, I think a 2 cent diode is preferable to a 5 cent capacitor since, (a) the capacitor still presents a low impedance or large inrush currents so you need to add another component resistor, and (b) over time electrolytics seem to fail more than diodes.
Check the power rating of your reed switch. Even though it says 0.5 Amps, you generally measure the switching power capability with the open circuit voltage. Small reed switches will be down there in the, say, 10 Watt region. So if you don't put in a resistor, you can easily be switching more than 1 Amp inrush to the capacitor...and with a track voltage of 18V you can be switching more than 10 Watts even though you would be carrying much less.
