Suppose $A$ has size $k$. For example, suppose $n=10$, $m=5$ and $k=4$. Then, a choice of $A$ is the same as diving up $\{1,2,\ldots,10\}$ into runs of 0s and 1s, with $k=4$ ones and no two ones adjacent. So, for instance, 0010101010.
The the $m$ intervals will contain a total of $n-k$ elements (the $0$s). The ways of choosing the lengths of the runs of $0$s correspond to compositions of $n-k$ into $m$ parts. There are $\binom{n-k-1}{m-1}$ ways to do that.
Similarly, there are $k$ $1$s, divided into $m-1$ runs, which is compositions of $k$ into $m-1$ parts, so $\binom{k-1}{m-2}$.
Once you have a composition for the $0$s and a composition for the $1$s, the set $A$ is totally determined (interleave the runs of $0$s and $1s$). So, summing over $k$ gives $\sum_{k=0}^n \binom{n-k-1}{m-1}\binom{k-1}{m-2}$