Here is my proof. After I show that the span(S) is a subspace of $V$, I am wondering if my proof is rigorous enough and doesn't have any holes on it. I feel okay about my reasoning, but not fully confident. Did I skip a step or assume too much in my leaps of logic?
Let the span(S) of any subset $S\subset V$ be denoted as span$(v_1, \dots v_m)$ for $v_i \in S$. First we need to check if the span(S) is a subspace of $V$.
0-vector: $0 = 0v_1 + \dots + 0v_m$ is in $S$
Closure under addition:
$$(a_1v_1 + \dots + a_mv_m) + (c_1v_1 + \dots + c_mv_m)\\ = (a_1+c_1)v_1 + \dots + (a_m+c_m)v_m$$
Closure under scalar multiplication for $\lambda \in \mathbb{F}$:
$$\lambda (a_1v_1 + \dots a_mv_m) = \lambda a_1v_1 + \dots + \lambda a_mv_m$$
Now by the given subspace properties of closure under addition and scalar multiplication, each $v_j$ is a linear combination of the span$(v_1, \dots , v_j)$.
Let $W$ be a subspace in $V$ that only contains a linear combination $x=a_1v_1 + \dots + a_mv_m$ for $v_i \in S$. Then $W$ is the smallest subspace of $V$ that contains $S$ because every vector $v_i \in S$ can be created from the linear combination.
Best Answer
The fact that $\text{span}(S)$ is a subspace of $V$ is OK. ($0$ being an element need not be verified, it follows by taking the scalar $0$.) Remark that it clearly contains $S$ as well (for any $s \in S$ take the trivial $1\cdot s \in \text{span}(S)$.
The proof that it is the smallest such: suppose that $W$ is any linear subspace of $V$ that contains $S$. Then any member $x \in \text{span}(S)$, i.e. any $x=\alpha_1 s_1 + \ldots \alpha_n s_n$, for some finitely many $s_i \in S$ and scalars $\alpha_i$, is in $W$ as all $s_i$ are ($S \subseteq W$ by assumption, and $W$ is closed under scalar multiplication and addition, being a subspace), so $\text{span}(S) \subseteq W$. This holds for any such $S$ and that is the definition of being the smallest subspace of $V$ that contains $S$.