Using Subset Sum algorithm $O(n)$ times to find the subset
Suppose that $\mathit{sum}$ is the sum of some subset of $A[2],\ldots,A[n]$. Then you can just remove $A[1]$.

Conversely, if $\mathit{sum}$ is not the some of any subset of $A[2],\ldots,A[n]$, then any subset summing to $\mathit{sum}$ must contain $A[1]$. Therefore it consists of $A[1]$ together with a subset of $A[2],\ldots,A[n]$ summing to $\mathit{sum}-A[1]$.

Either way, we have eliminated $A[1]$. We can then go on and eliminate $A[2],\ldots,A[n]$ in sequence.