Higman's group is not simple. Indeed, if you look at Higman's paper, you will see that his group is an amalgamated product of two groups $K_{1,2}=\langle a_1, a_2, b_2\mid a_1^{-1}a_2a_1=a_2^2, a_2^{-1}b_2a_2=b_2^2\rangle$ and $K_{3,4}=\langle a_3, a_4, b_4\mid a_3^{-1}a_4a_3=a_4^2, a_4^{-1}b_4a_4=b_4^2\rangle$ with free amalgamated subgroups $\langle a_1, b_2\rangle$ and $\langle a_3, b_4\rangle$ with $a_1=b_3, b_2=a_3$. Now take $K_{1,2}$ and add the relations $w(a_1,b_2)=1, w(b_2,a_1)=1$ for some "complicated" word $w$. You get a non-trivial group $K$. Similarly imposing $w(a_3,b_4)=1=w(b_4,a_3)$ on $K_{3,4}$ you get a non-trivial group $K'$. Then there is a homomorphism from the Higman group to the amalgamated product of $K$ and $K'$ with a non-trivial kernel. The fact that $K$ (and, equally, $K'$) is not trivial is not obvious but it is not very difficult to prove. 

<b> A simplification. </b> Instead of adding two relations, in fact it is enough to add a relation $w(a_1,b_2)=w(b_2,a_1)$ and similarly for $a_3,b_4$. For example, $(a_1b_2)^3=(b_2a_1)^3$. 

<b> Update 1. </b> Probably the easiest way to be convinced that $K$ ($=K'$) is non-trivial is to input its presentation into GAP or Magma (you do not need 3 in the simplified relation above, 2 is enough). 

<b> Update 2. </b> An even easier way is to impose further relations $a_2=b_2=1$ on $K$ (as in the simplification above). The factor-group is the infinite cyclic group. Hence $K$ is infinite (and $K'$ is infinite too). 

<b> Update 3. </b> I forgot to mention that one also needs that after we impose the relation as in the simplification, we should make sure that the subgroup of $K$ generated by $a_1,b_2$ has an automorphism switching $a_1$ and $b_2$ (otherwise we cannot form the proper amalgamated product of $K$ and $K'$). That is why we cannot replace 3 in the Simplification above by $1$.