If $I+J=R$, where $R$ is a commutative rng, prove that $IJ=I\cap J$. So I basically have to prove what is on the title. 


  Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $IJ=I\cap J$.


One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and $b_i\in J$. Thus for any fixed $i$, we have that since $a_i\in I$, we have that $a_ib_i\in I$, and the same argument shows that $a_ib_i\in J$, thus $\sum a_ib_i\in I$ and $\sum a_ib_i\in J$, this means that $x=\sum a_ib_i\in I\cap J$, and thus $IJ\subset I\cap J$. 

I am having troubles proving the other inclusion. Any comments?

Thanks
The statement you are trying to prove is only necessarily true for commutative rings with $1$. In this case, you can argue that
$$I\cap J\subseteq (I\cap J)R=(I\cap J)(I+J)=I(I\cap J)+ J(I\cap J)\subseteq IJ+ IJ=IJ$$

A counterexample to the statement for general rngs is given by endowing the group $\mathbb Z$ with the zero product, that is defining $a\cdot b=0,\forall a,b\in\mathbb Z$. The ideals $(2)$ and $(3)$ are still comaximal, as for any $x\in\mathbb Z$ we can write $x=a\times 2+b\times 3$ for some $a,b\in\mathbb Z$, where $\times$ denotes regular multiplication, and $a\times 2=2+\cdots+2\in(2)$ where addition is performed $a$ times, and similarly $b\times 3\in (3)$. But $(2)\cap (3)=(6)\neq (0)$, yet clearly $(2)(3)=(0)$.
The direct proof (for unital rings) also can be given. Assume $x\in I\cap J$. Consider such elements $i,i'\in I$ and $j,j'\in J$ that $x=i+j$ and $1=i'+j'$. Note that $i=x-j$, $x\in J$ gives $i\in J$ and $j=x-i$, $x\in I$ gives $j\in I$.

We have $x=i\cdot1+j\cdot1=i(i'+j')+j(i'+j')=ii'+ij'+ji'+jj'$. Using the commutativity we get $x=i'i+ij'+i'j+jj'$. So there are such elements $r_1,...,r_4\in R$, $i_1,...,i_4\in I$, $j_1,...,j_4\in J$ that $x=r_1i_1j_1+...+r_4i_4j_4$ (put $r_1=...=r_4=1$). So $x\in IJ$.