$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question.
Case 1 - if $R$ is infinite. Let $$R=R_1\cup R_2$$ be any decomposition of $R$ into two disjoint infinite subsets.
Clearly, we have $|R_1\cap M|\le 1$ for each $M\in\mc M$. Therefore the system $\mc M\cup\{R_1\}$ is almost disjoint and thus $R_1\in\mc M$ (from maximality).
But this means that the intersection $$R_1\cap R=R$$ is infinite — contradicting the "choosability".
Case 2 - if $R$ is finite and it has at least two elements. We can see relatively easily that $$\mathcal M'=\mathcal M\cup\{M\cup R; M\in\mathcal M\}$$ is almost disjoint. Maximality then implies that for each $M\in\mathcal M$ we have $$M\cup R\in\mathcal M.$$ Then we have $|(M\cup R)\cap R|\ge 2$, a contradiction.
Case 3 - if $R$ is a singleton. In this case the condition $|M\cap R|=1$ actually means $R\subseteq M$. So we get $\bigcap \mc M \ne \emptyset$.