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The empty subset always works: if S=∅, every vertex has degree 0, which is even. So the answer is at least 1.
Now weigh the two forces against each other:
If those n constraints were independent, you would expect 2m−n valid subsets. The answer is 2m−n+1, exactly twice that.
So one constraint must be redundant. Ask yourself: across the whole graph, what is always true about the sum of all vertex degrees?
The sum of all degrees is 2∣S∣, always even. So once n−1 vertices are even, the last one is even with no choice. One constraint comes for free.