**Claim.**$\newcommand{\mc}[1]{\mathcal{#1}}\newcommand{\sm}{\setminus}$
Any infinite [almost disjoint family](https://en.wikipedia.org/wiki/Almost_disjoint_family) $\mc A$ is uncountable.

*Proof.* Let $\mc A=\{A_n; n<\omega\}$ be a countable AD-family. By induction we define
\begin{align*}
b_1&=\min A_1\\
b_2&=\min \{k\in A_2\sm A_1; k>b_1\}\\
b_n&=\min \{k\in A_n\sm\bigcup\limits_{j=0}^{n-1}A_j; k>b_n\}\\
\end{align*}
Since for each $n$ the set $A_n\cap\left(\bigcup\limits_{j=0}^{n-1}A_j\right)$ is finite, the sets defined above are non-empty. (So there is a minimum.)

The set $B=\{b_n; n\in\omega\}$ is an infinite set which has a finite intersection with every set from$\mc A$. (Just notice that $B\cap A_n\subseteq \{b_1,b_2,\dots,b_n\})$. Therefore $\mc A$ is not maximal, as the set $B$ can be added without violating the almost disjointness.

**Corollary.** Any MAD-family is uncountable.

This basically says that for the almost disjoint number $\mathfrak a$ (defined as the smallest cardinality of a MAD-family on $\omega$) we have $\aleph_1\le \mathfrak a\le\mathfrak c$. This number is one of the [cardinal characteristic of the continuum](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum). (The name almost disjointness number is commonly used, too.)

**Corollary.** There is no MAD-family $\mc M$ with the property described in the question.

*Proof.* If we had $|R\cap M|=1$ for each $M\in \mc M$, this would give us a bijection between $R$ and $\mc M$. So $\mc M$ would be countable, a contradiction.