Let $S$ be a linearly independent set. Let $S'$ be a proper subset of $S$.
$Span$$(S') \neq Span(S)$
Let $v$ be an element of $S$ which is not contained in $S'$; such an element must exist because $S'$ is a proper subset of S. Since $v \in S$, we have $v \in$ $span$$(S)$. Now suppose that $v$ were also in $span$$(S')$. This would mean that there existed vectors $v_1, \ldots, v_n \in S'$ (which in particular were distinct from $v$) such that
$v = a_1v_1 + a_2v_2 + . . . + a_nv_n$,
or in other words
$(−1)v + a_1v_1 + a_2v_2 + . . . + a_nv_n = 0$.
But this is a non-trivial linear combination of vectors in S which sum to zero (it’s nontrivial because of the $−1$ coefficient of $v$). This contradicts the assumption that $S$ is linearly independent. Thus $v$ cannot possibly be in $span$$(S')$. But this means that $span$$(S')$ and $span$$(S)$ are different, and we are done.
So, we have $v = a_1v_1 + a_2v_2 + . . . + a_nv_n$. We can subtract $v$ from both sides of the equation and then argue that elements of $span$$(S')$ are linearly dependent.
Suppose a vector $v = a_1v_1 + a_2v_2 + . . . + a_nv_n$ is linearly independent. Can't we just subtract $v$ from both sides of the equation and say that the linear combination is linearly dependent every time?