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I have been trying to gain intuition on Ideals of Rings and came across this problem. I was hoping for some guidance in the right direction. Thank you.

To prove a non-empty subset $S subseteq R$ is an ideal, you need to show two things:

1. 
   $forall , a,b in S$, $a-b in S$.


2. 
   $forall r in R$, $rS=Sr=S$ (absorption property)

Let $B_1,B_2 in mathcal{M}_2(A)$. Then you need to demonstrate $B_1-B_2 in mathcal{M}_2(A)$.
$$begin{bmatrix}a&b\c&dend{bmatrix}-begin{bmatrix}p&q\r&send{bmatrix}=begin{bmatrix}a-p&b-q\c-r&d-send{bmatrix}.$$
Since $A$ is an ideal, so for $a,p in A$, we will have $a-p in A$. Similarly you can deal with remaining entries. So the final matrix is in $mathcal{M}_2(A)$.

For the absorption property:

Let $J=begin{bmatrix}x&y\z&wend{bmatrix} in mathcal{M}_2(R)$. You need to show that $J[mathcal{M}_2(A)]=mathcal{M}_2(A)=[mathcal{M}_2(A)]J$. Said differently, you need to show that for all $B_1 in mathcal{M}_2(A)$, $JB_1 in mathcal{M}_2(A)$ and $B_1J in mathcal{M}_2(A)$.
$$JB_1=begin{bmatrix}x&y\z&wend{bmatrix}begin{bmatrix}a&b\c&dend{bmatrix}=begin{bmatrix}xa+yc&star\star&starend{bmatrix}.$$
Since $A$ is an ideal and $a,c in A$, so $xa, yc in A$ (by absorption property). But $A$ is a subring so closed under addition, which implies $xa+yc in A$. Now try doing this for remaining entries.

See if you can show the other containment.