Theorem: Let S be a linearly independent set in a vector space. If $\vec{X} \in V$ and $\vec{X} \in span \left(S\right)$ then $S \cup \{\vec{X}\}$ is linearly independent.
Proof:
It suffices to show, by contra positivity, that $\vec{X}$ $\in$ $span\left(S\right)$.
Let $S = \{\vec{v}_{i}\}_{i=1}^{n}$
$\vec{0} = c_{X}\vec{X} + \sum_{i=1}^{n}c_{i}\vec{v}_{i}$ with coefficients $c_{i} = c_{X} = 0, \forall i \in \mathbb{Z}_{1}^{n}$.
Rearranging:
$\vec{X} = \frac{1}{c_{X}} \sum_{i=1}^{n}c_{i}\vec{v}_{i}$
Here, I am unable to move forward. The reciprocal of $c_{X}$ gives an undefined value.
Any help to move this proof forward is appreciated. Thanks in advance.
You want $\vec{x} \notin \operatorname{span}(S)$, as the assumtion of course (as you do have in the title).
In that case we can show that $S \cup \{\vec{x}\}$ is lin. independent:
Suppose no, then there are vectors $\vec{s_1},\dots \vec{s_n} \in S$ and coefficients $c_1,\ldots c_n, c$, not all $0$, such that
$$\sum_{i=1}^n c_i \vec{s_i} + c\vec{x} = 0$$
If $c =0$ we have an immediate contradiction with the fact that $S$ is linearly independent, because not all the $c_i$ are $0$.
So we must have $c \neq 0$ and so $\frac{1}{c}$ is a well-defined scalar.
But then $$\vec{x} = \sum_{i=1}^n \frac{1}{c}(-c_i) s_i$$ and we'd have that $\vec{x} \in \operatorname{span}(S)$ contrary to the (corrected) assumption.