a) Let $a$ be a member of $D$ and let $x$ be a member of $A$. If $x$ is invertible, then $ax$ is a member of $D$, so $axax = 0$, and $axa = 0$. If $x$ is a member of $D$, then $1+x$ is invertible, so $a + ax = a(1+x)$ is also a member of $D$. Hence $0 = (a+ax)^2 = a^2 + a^2x + axa + axax = axa(1+x)$, and consequently $axa = 0$.

b) Let $P$ be the set of all finite non-zero products of elements of $D$. Since $|D| \ge 2$, the set $P$ is non-empty. Notice that if $a_1a_2\cdots a_k$ is a product in $P$, then $a_i \ne a_j$ for $i \ne j$. Indeed, if $a_i = a_j$ for some $i < j$, then $a_1a_2\cdots a_k = a_1a_2\cdots a_i(a_{i+1}\cdots a_{j-1})a_i\cdots a_k = 0$, by a), which is a contradiction. Hence $P$ is a

finite set. Finally, let $a = a_1a_2 \cdots a_k$ be a product in $P$ of maximal length $k$, and let $b$ be a member of $D$. If $b$ is one of the factors of $a$, then $ab = ba = 0$, by a). Otherwise, the words $ab$ and $ba$ both have length greater than $k$, so they cannot belong to $P$, by maximality of $k$, and again $ab = ba = 0$.