Independent Set Problem Variant, Induction
Hint:  Suppose you have a graph $G = (V,E)$ which is yes, and then there is a vertex $v \in V$ such that $G - v$ (the graph where you have removed $v$ from the graph) is no.  What can you say about $v$ and $G$?

Let us take an example, namely the following graph:

$G= (\{a,b,c,d\}, E=\{ab, bc, cd, ad, ac\})$

a —— d
| \  |
|  \ |
b —— c


and with $k=2$.  This is currently a yes instance.

First we ask:  If we delete $a$ from $G$, will the resulting instance still be a yes instance?

     d
     |
     |
b —— c


The answer is yes, this graph is still a yes instance.

Then we delete $b$ and ask the same question:

     d
     |
     |
     c


The answer is now no, this instance is no longer a yes instance.  Hence we cannot delete $b$.  Instead we try to delete $c$:

     d


b     


The answer is still yes, and $|V(G')| = k$, so we conclude that $\{b, d\}$ is an independent set in $G$ of size $k$.