Here's a combinatorial proof. Suppose [tex]A[/tex] has [tex]n[/tex] elements.
For a subset to contain [tex]a[/tex], it must consist of at least one element. So if any given subset has [tex]k[/tex] elements, where [tex]1\le k\le n[/tex], then [tex]a[/tex] is not one of the other [tex]k-1[/tex] elements. This means the number of subsets containing [tex]a[/tex] is
[tex]\displaystyle\sum_{k=1}^n\binom11\binom{n-1}{k-1}[/tex]
Put another way, we are choosing elements from [tex]A[/tex] to form a subset of [tex]k[/tex] elements. We want [tex]a[/tex] to be in each subset, so we have [tex]n-1[/tex] other elements of [tex]A[/tex] from which to choose. Then we sum over all the possible sizes of the desired subset.
On the other hand, if we want to build subsets not containing [tex]a[/tex], then we have [tex]n-1[/tex] total elements to choose from, and we can make subsets of size ranging from 0 to [tex]n-1[/tex], so the number of subsets not containing [tex]a[/tex] is
[tex]\displaystyle\sum_{k=0}^{n-1}\binom10\binom{n-1}k[/tex]
We have [tex]\dbinom10=\dbinom11=1[/tex], and in the second sum we can shift the index up by 1 to get
[tex]\displaystyle\sum_{k=1}^{n-1+1}\binom10\binom{n-1}{k-1}[/tex]
which is the same as the first count.
