Linearly Independent Subset also Independent in Generated Subspace

Theorem
Let $$G$$ be a finitely generated $K$-vector space.

Let $$S$$ be a linearly independent subset of $$G$$.

Let $$M$$ be the subspace of $$G$$ generated by $$S$$.

If $$M \ne G$$, then $$\forall b \in G: b \notin M$$, the set $$S \cup \left\{{b}\right\}$$ is linearly independent.

Proof
Suppose that $$\sum_{k=1}^n \lambda_k x_k + \lambda b = 0$$, where $$\left \langle {x_n} \right \rangle$$ is a sequence of distinct vectors of $$S$$.

If $$\lambda \ne 0$$, then $$b = - \lambda^{-1} \left({\sum_{k=1}^n \lambda_k x_k}\right) \in M$$ which contradicts the definition of $$b$$.

Hence $$\lambda = 0$$, and so $$\sum_{k=1}^n \lambda_k x_k = 0$$.

Therefore $$\lambda_1 = \lambda_2 = \cdots = \lambda_n = \lambda = 0$$ as $$S$$ is linearly independent.