Proving there is no polynomial algorithm for independent set I need some guidance in an assignment I'm doing.

I'm at complete loss, he says the the MAXIMUM INDEPENDENT SET problem is NP-hard and then asks me to prove that there is no polynomial time for the same algorithm and it also guarantees that some inequality holds, which I also don't understand, since $A(G)$ returns the size of a maximal independent set in $G$ and $OBT(G)$ returns the size of an independent set of maximal size for $G$ but isn't the case that a graph could have multiple maximal independent sets? Let's assume that it returns the largest one which reduces the epression to $0 \le K$ which should always holds since $K$ is a natural number?

As for the hint given, I don't see how we can use $H$ if I understand it correctly, it says split $G$ into disjoint sets. Also it says for every instance $G$ which is confusing me.

When I asked the professor she was very reluctant to help me, so please any guidance is appreciated. 

The full question:   .
Suppose that there exists an integer $K$ and a polynomial time algorithm $A$ which, when run on a graph $G$, outputs a value $A(G)$ which satisfies
$$
 |A(G) - \alpha(G)| \leq K,
$$
where $\alpha(G)$ is the maximum size of an independent set in $G$.

We will show that $A$ can be used to determine $\alpha(G)$ in polynomial time, which contradicts the assumption $\mathsf{P} \neq \mathsf{NP}$.

Given a graph $G$, let $H$ consist of $2K+1$ disjoint copies of $G$; note that $\alpha(H) = (2K+1) \alpha(G)$. Run $A$ on $H$ to get a value $A(H)$ which satisfies
$$
 |A(H) - \alpha(H)| \leq K \Longrightarrow |A(H) - (2K+1) \alpha(G)| \leq K \Longrightarrow \\
 (2K+1)\alpha(G) \in \{A(H)-K,\ldots,A(H)+K\}.
$$
The set on the right contains $2K+1$ values, exactly one of which is a multiple of $2K+1$, hence we can determine $\alpha(G)$ given $A(H)$.