Let $A_n$ be the number of ways of partitioning an $n$-element set $M$ into $2$-element subsets. Pick $x \in M$. There are $n - 1$ choices for the element that is going to belong to the same partition as $x$, and then the remaining $n - 2$ elements can be partitioned into $2$-element subsets in $A_{n - 2}$ ways. Clearly $A_1 = 0$ and $A_2 = 1$, so we have:
\begin{align}
A_n &= (n - 1)A_{n-2}\\
A_2 &= 1\\
A_1 &= 0
\end{align}
Hence $A_n = 0$ for odd $n$ and for even $n$ we have:
$$
A_{n} = 1 \cdot 3 \cdot \ldots \cdot n -1.
$$
It is trivial to show the statement for $|a|\ge1$. W.L.O.G we can say that $0<a<1$.
Now, let's suppose that sets $A$ and $B$ s.t. they satisfy the given conditions exist.
1. case
$a< \frac{1}{2}$
Considering that $A$ and $B$ are disjoint, the fact that $a>0$ and that element $1$ has to be in one of them it must be $1\in B$, this means that $\max A=1-a$. Similarly $\min B=a$. This implies, not only $[0,a)\subset A$, but also $2a\in A$. Continuing like this we can see that all subsets of $[0,1]$ of the form $[2ka,(2k+1)a)$ are subsets of $A$ and all subsets of the form $[(2k+1)a,(2k+2)a)$ are subsets of $B$. Keep in mind that $k\in \mathbb{N}$, until $2ka=1$. This means that there exists some $n\in \mathbb{N}$, s.t. $2na=1$, which would mean $1\in A$, but of course that is impossible, therefore a contradiction.
2. case
$a>\frac{1}{2}$
You can simply show that both $A$ and $B$ in this case would have to be intervals, since $\min B=a>\frac{1}{2}$ and $2a\notin A$, because if it were that $A$ wouldn't be a subset of $[0,1]$. Because $max A=1-a$, that leaves $(1-a,a)\not\subset A\cup B$. So obviously $A\cup B \neq [0,1]$. Contradiction!
3. case
$a=\frac{1}{2}$
In this case, $\max A=\min B=\frac{1}{2}$. Implying $A \cap B=\{\frac{1}{2}\}$. And again, a contradiction.
From all of this we can conclude that such sets $A$ and $B$ that satisfy the given conditions don't exist.
Best Answer
Let $S_i$ for $i=1,2,\ldots,N$ denote the $N$ given subsets of $S$. Suppose that $X$ is a subset of $S$ uniformly randomly chosen from all the $2^n$ subsets of $S$. Define $Y(X):=S\setminus X$.
For a fixed value $i\in\{1,2,\ldots,N\}$, the probability that $S_i\subseteq X$ is $\dfrac{2^{n-m}}{2^n}=\dfrac{1}{2^{m}}$ (because $|S_i|=m$ and $X\setminus S_i$ is a random subset of $S\setminus S_i$). Similarly, the probability that $S_i\subseteq Y(X)$ is $\dfrac{1}{2^m}$. Therefore, the probability $p_i$ that $S_i\subseteq X$ or $S_i\subseteq Y(X)$ is $$p_i\leq \frac{1}{2^{m}}+\frac{1}{2^{m}}=\frac{1}{2^{m-1}}\,.$$ (For a positive integer $m$, $p_i=\dfrac1{2^{m-1}}$, but I just want to include the trivial case $m=0$. The OP has never said that $m$ has to be a positive integer.)
Since $N<2^{m-1}$, we obtain $$\sum_{i=1}^N\,p_i\leq \frac{N}{2^{m-1}}<1\,.$$ However, the probability that $S_i\subseteq X$ or $S_i\subseteq Y(X)$ for some $i\in\{1,2,\ldots,N\}$ is at most $\sum\limits_{i=1}^N\,p_i$ (due to subadditivity of probability measures). Consequently, the probability that $S_i\subseteq X$ or $S_i\subseteq Y(X)$ for no $i\in\{1,2,\ldots,N\}$ is at least $$1-\sum\limits_{i=1}^N\,p_i>0\,.$$ Ergo, there exists $A\subseteq S$ such that there are no $i\in \{1,2,\ldots,N\}$ for which $S_i\subseteq A$ or $S_i\subseteq Y(A)=:B$.