What should I expect the low end of my multimeters resistance measurement range to be?
Summary


A method is given to allow use of a voltmeter and a known higher value resistor. This will give better results than available ohms range will. 
Probe resistance and contact resistance need to be allowed for if using meter.
Meter errors on 400 ohm range are in excess of value to be measured but a 'rough idea' can be established. 




What Chris says, plus:


Rated accuracy is with respect to full scale :-( 

= about 1.5 ohms on 400 ohms notional 

AND +/- 2 digits LSD flicker as well. 
Twist plugs to and fro in sockets to make good contact. 
Short lead tips together - rub against each other and note lowest reading. 
THEN measure component - rub probe tips against component or lowest reading. This - probe tip shortest version should give ~= OK result. 


BUT at 0.22 ohms you want a better method. Chris's series current method is good. If you don't have a variable current supply use variable voltage + larger resistor or even a battery and resistor.

eg to get 0.22 ohms to give 0.22 Volt say you need 1 Amp - probably too high. To get 22 mV you need 100 mA. To get 2.2 mV (may be OK if you have a 200 mV range) you need 10 mA. A 9V battery and a 1K resistor gives 9 mA.

One method using voltmeter and known larger value resistor.

This works by having same current in a known large resistor and the target resistor. The voltage drops are proportional to the resistances. The example below uses a current of ABOUT 10 mA for a resolution of 10 mV per ohm. Using a 100 R resistor in place of the 1K below gives a resolution of 100 mV/ohm. This allows an eg nominally 9V battery and a known resistor to be used. Exact battery voltage does not matter as long as it does not "droop" (change) during measurement. Accuracy of measurement depends on voltmeter accuracy and resistor accuracy. Be aware of the possible effects of resistor heating. At 1k and 9V power = V^2/R = 0.081 W. At 100R and 9V power = 0.81 Watt! - change of resistance with heating may be significant depending on the resistor used. 


Measure 1k resistor as accurately as you can = R_1k. Say you get 975 ohms for examples sake. 
Place 1k + 0.22R in series with 9V battery. 
Measure voltage across 1k - say 8.85 V as an example.
Measure V across 0.22R - say 1.7 mV as an example.


Then R_0.22 / R_1k = V_0.22 / V_1k

So R_0.22 = R1k x V_0.22 / v1k

Here R_0.22 = 975 x 0.0017 / 8.85 = 0.187 ohms.