Suppose $V$ is a vector space and that $X\subset V$ is a linearly independent subset of $V$. If $S(X)$ is a proper subset of $V$, then there exists a set $X’\subset V$ such that $X$ is a proper subset of $X′$, and $X′$ is a linearly independent set.
How do I prove this claim?
I assume $S(X)$ refers to $\operatorname{span}(X)$, and I will refer to the field we are working over as $F$.
Since $\operatorname{span}(X)\ne V$, there exists a vector $a\in V\setminus \operatorname{span}(X)$. I claim that $X’ = X\cup\{a\}$ is linearly independent. Because $X$ is already linearly independent, it suffices to just check linear combinations containing the newly added element $a$.
Suppose $c_1x_1 + \ldots + c_nx_n + c_{n+1}a = 0$, where each $c_i\in F$ and $x_i\in X$. Because $F$ is a field, each $c_i\ne 0$ has a multiplicative inverse $c_i^{-1}$, whence
$$c_{n+1}a = -(c_1x_1+\ldots + c_nx_n)$$
and thus, if $c_{n+1}\ne 0$,
$$a = -c_{n+1}^{-1}(c_1x_1+\ldots + c_nx_n)\in\operatorname{span}(X).$$
So, if $c_{n+1}\ne 0$, $a\in\operatorname{span}(X)$, which is a contradiction. As such, $c_{n+1}=0$. But then we have
$$c_1x_1 + \ldots + c_nx_n=0,$$
which means each $c_i=0$ due to the linear independence of $X$. Therefore, $X’$ is a linearly independent set.