Solution:
Consider a table with rows corresponding to the elements of $A$ and columns corresponding to the elements of $M$. We write $\times$ in a cell if the element of $A$ in the respective row is adjacent to the element of $M$ in the respective column. Let $|M| = k$. It follows from the given condition that there are no two equal rows, which means that $M$ has at least $16$ different subsets. Since a set with $n$ elements has $2^{n}$ subsets, we get $k \geq 4$.

Any row has exactly five adjacent and thus any column contains exactly five $\times$, i.e., the total number of $\times$ is $5k$. A minimal number of $\times$ by rows is attained if one row contains no $\times$, $k$ rows contain one $\times$, $\binom{k}{2}$ rows contain two $\times$, etc.

If $k = 4$, then all the subsets of $M$ are $16$ and hence any subset of $M$ appears exactly once as an adjacent set to some element of $A$. Then we have one row with no $\times$, $4$ rows with one $\times$, $6$ rows with two $\times$, $4$ rows with three $\times$ and one row with four $\times$. The total number of $\times$ becomes $32 > 20$, a contradiction.

For $k = 5$ one has $25$ $\times$. Their minimal number by rows is attained when one row contains no $\times$, $5$ rows contain one $\times$ and $10$ rows contain two $\times$. Since $5 \cdot 1 + 10 \cdot 2 = 25$ the distribution of $\times$ must be exactly the one described above. This means that any two elements of $M$ are simultaneously adjacent to some element of $A$.

It is easy to see that if two elements of $M$ do not coincide at at most two positions, then there is no element of $A$ that is adjacent to them. Hence any two elements of $M$ do not coincide at one or two positions. If there are two elements which do not coincide at one position, we may assume that they are $a = 0000$ and $b = 1000$. The adjacent to $a$ and $b$ in $M$ are $0000$ and $1000$. So the rows of $a$ and $b$ coincide, a contradiction. This shows that any two elements of $M$ are different at exactly two positions. We may assume that $0000 \in M$. Then the remaining $4$ elements are among $0011$, $1100$, $0101$, $1010$, $1001$ and $0110$. But among any of the pairs $(0011, 1100)$, $(0101, 1010)$ and $(1001, 0110)$ at most one element can be chosen, a contradiction.

For $k = 6$ the set $M = \{0000, 1111, 0111, 0100, 1001, 0101\}$ separates any two elements of $A$.

Hence the desired minimal number is $6$.