Let $A$ be a commutative graded ring, $I\subset A$ a homogenous ideal, and $J\subset A$ a non homogenous ideal. Is $I\cap J$ homogenous?
I feel like it should be. Indeed consider the following proof:
Let $i\in I\cap J$, then $i\in I$ and $i\ in A$. Since $i\in I$, and $I$ is homogenous it follows that: \begin{align} i=\sum_i a_ib_i \end{align} where $a_i\in A$ and $b_i\in I\cap A_i$. Since any element can be written in $I\cap J$ can be written as a linear combination of homogenous it follows that $I\cap J$ is generated by homogenous elements and is thus homogenous.
Is this the correct way of thinking about this? It feels a little too easy...
It's not correct. Note that a generating set by definition has to be a subset of the ideal. In your case there is no good reason why the elements $b_i$ should be in $I\cap J$, and so you can't claim they generate this ideal. Actually, in a graded ring any element is a sum of homogeneous elements, but it doesn't mean every ideal is homogeneous.
For an easy counterexample to your claim, take $A=\mathbb{C}[x]$. The non homogeneous ideal $(x^2+x)$ is contained in the homogeneous ideal $(x)$, and so their intersection is non homogeneous.