Reduction from Independent Set with fixed vertex to Independent Set
As I understand, your problem is a decision problem defined as such:
Independant set with fixed vertex (ISFV):

Input: a graph $G = (V, E)$, a vertex $u \in V$, an integer $k$.
Question: is there an independent set of size $k$ containing $u$?

Independent set (IS) is defined as:

Input: a graph $G = (V, E)$, an integer $k$.
Question: is there an independent set of size $k$?

Suppose you can solve ISFV. Then you can solve IS by running ISFV for each $u\in V$ and checking if the answer is yes for any $u\in V$. Since there are a polynomial number of vertices, the reduction is indeed polynomial.
Another way to do it is to construct the graph $G' = (V\cup\{u\}, E)$ (adding a vertex with no other edge), and check ISFV with $G'$, $u$ and $k + 1$, since the vertex $u$ can always be added to an independent set.