Let $R$ be an infinite commutative ring. Which of following options is false?
Center of $M_2(R×R)$ is nontrivial.
$ M_2(R×R) \cong M_2(R)×M_2(R)$
The number of units in $M_2(R ×R)$ is infinite.
The number of two-sided ideals in $M_2(R ×R)$ maybe are finite.
Now "1" is true because every diagonal matrix is in center of this ring.
"4" is true because :
Theorem: every two-sided ideal of $M_n(R)$ has the form $M_n(I)$ for some unique two-sided ideal $I$ of $R$.
Your reasoning is correct.
Your argument is not quite correct. One can use the following, for a commutative ring $R$ with $1$ and $A$ an $R$-algebra -
$$M_n(R) \otimes A \simeq M_n A$$
Note that the above is in fact an $R$-algebra isomorphism. Now let $A = R^2$ and use the commutativity of tensor products with direct sums.
Coupled with (2), this follows from the fact that $M_2 R$ has infinitely many units provided $R$ is infinite. Note that this does not depend upon the number of units in $R$.
This is true. In fact one can prove that $M_2(\mathbb{F}^2)$ has only 3 proper ideals for any field $\mathbb{F}$.