If you have 5V dc, you can supply its input LED with a current of 5mA by adding a resistor (Rs), in series between Vcc and input:
Rs = ( Vcc - V_led ) / I_led
Rs = ( 5 - 1.2 ) / 5 = 0.76 K
Rs=680R could be used (I_led would be a bit higher than 5mA).
Using an ohmmeter, check if the resistance of its output is high (as being open) before applying the 5mA current.
While the probes of the meter are connected, the measured resistance should become relatively low (not necessarily very low) after applying the input current.
Please note that even if you remove the input current the state of the output (conductive) won't change since its ouput is a triac.
The triac turns off only if the output current becomes lower than its holding current (from datasheet, it is typically 100uA).
These method may not work if the current provided by the ohmmeter is very low (lower than 100uA).
In some multimeters, the diode checker option (position) usually supplies, via the probes, a relatively high current (not less than 1 mA in general).