The circuit is this:
The diodes are ideal. The solution for this question is D1 OFF and D2 ON, and hence current \$I = 3mA\$.
But when I assume that both D1 and D2 are OFF, I am not able to prove that my assumption is wrong.
How can I analyze the circuit if I assume that both diodes are OFF?
Just for the benefit of the larger audience: when you deal with purely resistive circuits containing n ideal diodes you can analyze them by guessing the state of every diode (either ON or OFF), do the calculations and then check if the results are coherent with the assumption you have made.
If you guessed right, i.e. the calculations confirm the assumption, then it can be shown that you've found the solution, which is unique. Otherwise you have to redo all the calculations changing your assumption about the state of the diodes. At worst you have to check \$2^n\$ circuit configurations (all the possible combinations of states of the n diodes), but with a bit of common sense and experience you can make the right guess with many fewer attempts.
Ok, then, but how do you check whether your assumptions are correct? Just remember what's the behavior of an ideal diode in either of its two states:
So, let's consider your specific circuit. If you assume both diodes are OFF, you replace them with open circuits. Then think what is the voltage at the upper leg of the resistor: since no current flows in the resistor there is no voltage drop across it... you should be able to continue from here and see that the results contradict the assumption. Hence the diodes cannot be both off.
Very late answer. My goal is just to explicitly answer the question, for future readers who might have not understood Lorenzo's answer (read it first!)
For an ideal diode, the \$i\$-\$v\$ characteristics is divided into two cases, depending whether the diode is ON or OFF. In the following equations, the reference direction of the current and the reference polarity of the voltage are the typical ones, i.e. the arrow of \$i\$ points in the direction of the diode's symbol arrow, and the \$v\$ is defined according to the passive sign convention (\$+\$ sign where \$i\$ enters). If a diode is OFF, then
\$ i = 0, v < 0 \tag*{} \$
and if it's ON, then
\$ v = 0, i > 0 \tag*{} \$
So if you assume both diodes are off, then their voltages must be negative. If you get a positive voltage, the assumption is false. After replacing both diodes with an open circuit and applying KVL, you get that
\$ -1 + v_{D1} - 3 = 0 \implies v_{D1} = +4 \text{ V} > 0 \tag*{} \$
and
\$ -3 + v_{D2} - 3 = 0 \implies v_{D2} = +6 \text{ V} > 0 \tag*{} \$
Since their voltages are not negative, you got a contradiction. So they can't be OFF.
