Well, strictly speaking, this is only true for $k > 0$. For $k=0$ we have a graph with no edges, and the probability of picking an independent set is equal to $1=\frac{1}{2^0}$.
As for your approach, I didn't really understand what it is. If you consider any given selection of vertices, it is either independent or not. It is not a matter of probability, because the graph is not random. The graph is fixed, the selection is random.
Here is a possible way to approach this. For any fixed selection of vertices, the probability of picking this exact selection is $\frac{1}{2^n}$, where $n$ is the number of vertices in the graph. Here I use the assumption that we pick different vertices independently from each other. It is probably implied.
Now, from this observation we can see that the probability of picking an independent set of vertices is simply $ p = \frac{m}{2^n}, $ where $m$ the total number of different independent sets of vertices.
So, it comes down to proving that $m>2^{n-k}$. It isn't hard. Here is a hint that might help with that: you can use the fact that a graph with $n$ vertices and $k$ edges has at least $n-k$ connected components.