Let $R$ be a ring and $M$ be an irreducible $R$ module. I want to create an irreducible $A$-module, where
$$A = \begin{pmatrix}R & R\\0 & R\end{pmatrix}.$$
I defined $\bar{M}$ as $$\bar{M} = \{(m, 0) : m \in M\}.$$
$\bar{M}$ is an $A$-module in the natural way. So let $J$ be a $A$-submodule of $\bar{M}$. Then, $$J\begin{pmatrix}R & R\\0 & R\end{pmatrix} \subseteq J.$$
Let $p\in J$. Since $J\subset\bar{M}$, $p=(m,0)$, where $m \in M$. Thus $$p\begin{pmatrix}R & R\\0 & R\end{pmatrix} = (m,0)\begin{pmatrix}R & R\\0 & R\end{pmatrix} = (mR,0).$$
Now, since $M$ is an irreducible $R$-module, and $m\not=0$, $mR=M$ (is it true that $mR\not=(0)$?). We show that $$p\begin{pmatrix}R & R\\0 & R\end{pmatrix} = (M,0) = \bar{M}.$$
This implies that $$\bar{M} = p\begin{pmatrix}R & R\\0 & R\end{pmatrix} \subseteq J \subseteq \bar{M} \implies J = \bar{M}.$$
So, $\bar{M}$ is an irreducible $A$-module. Is this proof correct?
Your proof is correct, except for one small mistake.
If $M \times M$ is considered as a right $A$-module using the obvious definition of scalar multiplication (i.e., $\begin{pmatrix}m&n\end{pmatrix} \begin{pmatrix}a&b\\0&c\end{pmatrix}=\begin{pmatrix}ma&mb+nc\end{pmatrix}$), then $M \times \{0\}$ is not a submodule, but $\{0\} \times M$ is a submodule. So, $\bar{M}$ should actually be defined as $\{0\} \times M$, not as $M \times \{0\}$.
Is it true that $mR \ne (0)$? Of course, it is true (assuming that $R$ is a unital ring and $M$ is a unital right $R$-module), since $0 \ne m = m1 \in mR$.
If rings and modules are not required to be unital, then $mR$ could be zero for some nonzero $m \in M$. For example, $\mathbb{Z}/p\mathbb{Z}$ with the trivial zero scalar multiplication would be an irreducible $R$-module for any prime $p$, and then $mR=(0)$ for every $m \in \mathbb{Z}/p\mathbb{Z}$.
Other than $\mathbb{Z}/p\mathbb{Z}$ for prime $p$ with the trivial zero scalar multiplication, there are no other counterexamples (up to isomorphism) in the nonunital case. Indeed, $\{m \in M : mR = (0)\}$ is easily seen to be a submodule of $M$, and so by irreducibility, it must be either $(0)$ or $M$. If it is $M$, then $M$ would have the trivial zero scalar multiplication and the submodules of $M$ would then just be the additive subgroups. The irreducible abelian groups (or $\mathbb{Z}$-modules) up to isomorphism are known to be just $\mathbb{Z}/p\mathbb{Z}$ for prime $p$, so $M$ must be isomorphic to one of them.
In the $\mathbb{Z}/p\mathbb{Z}$ case, $\bar{M}$ would still be isomorphic to $\mathbb{Z}/p\mathbb{Z}$ with the trivial zero scalar multiplication, so it would still be an irreducible $A$-module.