Let $k$ be a field. Consider the ideals $I_1=(x),I_2=(y),J=(x^2,y)$ of $R=k[x,y]/(xy,y^2)$. Show that the homogeneous elements of $J$ are contained in $I_1\cup I_2$ but that $J\not\subset I_1$ and $J\not\subset I_2$.
First of all, what is meant by "homogeneous elements of $J$"? The notion of homogeneous element makes sense in a graded module $M=\oplus_{i=-\infty}^{\infty}M_i$ over a graded ring $R=R_0\oplus R_1\oplus\dots$. The ideal $J$ is an $R$-module, but how is the grading of $J$ and $R$ defined?
Also, whichever the definition of a homogeneous element is, I think the element $x+y$ must be homogeneous according to the definition. But it doesn't lie in the union $I_1\cup I_2$...
The grading on $k[x,y]$ induces a grading on the quotient $R=k[x,y]/(xy,y^2)$; the abelian groups $$A_n:=\bigoplus_{i=0}^nx^iy^{n-i}k,$$ give the grading on $k[x,y]=\bigoplus_{n\geq0}A_n$. Let $B_n$ be the image of $A_n$ in the quotient $R$, so that $$B_0=k,\qquad B_1=\overline{x}k\oplus\overline{y}k \qquad\text{ and }\qquad B_n=\overline{x}^nk \text{ for }n>1.$$ Then $R=\bigoplus_{n\geq0}B_n$ is a graded ring. Because $J$ is an $R-$submodule of $R$, this gives a grading on $J$ as well.
As for your suspected counterexample to the claim; note that $x+y$ doesn't lie in $J$ either.