How to prove Q1 and Q2 are equal using state equivalence algorithm
The algorithm is working fine on this DFA. At step 2 you'll get the table as:

  1 2
1 G G
2   G


G is green.

If there is an unmarked pair (Qi, Qj), mark it if the pair {δ(Qi, A), δ (Qj, A)} is marked for some input alphabet. Where A is set of all inputs.

{δ(1, 'a'), δ(2, 'a')} = {2,1} = {1,2}
{δ(1, 'b'), δ(2, 'b')} = {1,2}

{1,2} or {2,1} is not marked. Hence move to step 4, Combine all the unmarked pair (Qi, Qj) and make them a single state in the reduced DFA.

I prefer the method of partitioning for minimizing the DFA. Consider the following DFA 
State transition table for the DFA will be:

   0  1
a  b  c
b  a  d
c  e  f
d  e  f
e  e  f
f  f  f


Draw 0 equivalent partitions, by separating final and non-final states in different sets.
[a b f] [c d e]
Draw 1 equivalent partitions, by taking every possible pair from each set and checking whether their transitions on an input symbol belongs to a single set or not. For example {a,b} on 0 goes to {a,b} and it belongs to a single set [a b f], {a,b} on 1 goes to {c,d} which belongs to 1 single set [c d e], hence {a,b} are 1 equivalent. But {a,f} on 1 goes to {c,f} and it doesn't belong to a single set, hence they are not equivalent. So separate them. (you can check for {b,f},{c,d},{c,e},{d,e} as well, result will be the same)
[a b] [f] [c d e]
Draw 2 equivalent partitions, by repeating the procedure and you'll find nothing is changed. Hence final configuration is reached.
Merge [a b] and [c d e]



In your example: 0 equivalent partitions: [q1 q2] 1 equivalent partitions: [q1 q2]

Hence they are combined.