Basis on vector space $V$ If $S_i$ is a set of linearly independent vectors of vector space $V$ and $S_g$ a set of generators of $V$. Prove that it exist $S'_g\subset S_g$ that $S_i\cup S'_g$ is a basis of $V$.

Notice that $V$ is not necessarily finite dimensional.
If you are asking about finitely-dimensional spaces, this is exactly Steinitz exchange lemma.

If you are interested in infinitely-dimensional space, I think that you only need to modify a little the standard proof of existence of Hamel basis based on Zorn Lemma. This proof can be found in many places. (Or maybe I should say that it is more similar to the proof of the fact, that every linearly independent subset of $V$ is contained in a Hamel basis of $V$.)

We will work with the system
$$\mathcal P=\{S\subseteq S_g; S\cup S_i\text{ is linearly independent}\}.$$
We check that for any chain $\mathcal C\subseteq\mathcal P$ we also have $\bigcup\mathcal C\in\mathcal P$. So the poset $(\mathcal P,\subseteq)$ fulfills the assumptions of Zorn Lemma.

By Zorn Lemma there exists a maximal $S'_g\in\mathcal P$. It only remains to show that $S'_g\cup S_i$ generates $V$. (We will use maximality of $S'_g$ in the proof.)
Say $S_i = \{\vec v_1, \vec v_2, ... \vec v_n \}$.   

Assume that there does not exist any vector $\vec v \in S_g$ such that $\{ \vec v \cup S_i \}$ is linearly independent.
  Now for every vector $\vec v \in S_g$ there exists $a, a_1, a_2, ...$ such that $ a\vec v + a_1 \vec v_1 + a_2 \vec v_2 + ... + a_n \vec v_n = \vec 0 $.
  However, by the linear independence of $S_i$,  $a \neq 0$.
  Now we have a expression of $\vec v$ in terms of $ \vec v_1, \vec v_2, ... \vec v_n $, and so we have shown $S_i$ itself generates $V$.  

Else, there exists a $\vec v \in S_g$ such that $\{ \vec v \cup S_i \}$ is linearly independent. Then just add this $\vec v$ to $S_i$.  

By repeating this procedure, we will end up with a linearly independent extension of $S_i$ that generates $V$, because $S_g$ generates $V$ so even in the worst case, we can add all of $S_g$ to $S_i$.  

The vectors of $S_g$ we have added to $S_i$ forms the desired subset $S_g' \subseteq S_g$.