Proofs from Spans

For one vector, $\lambda_1v_1\in V$ by scalar closure. Suppose every combination of $k$ vectors lies in $V$. If

then scalar closure gives $\lambda_{k+1}v_{k+1}\in V$ and addition closure gives

Induction proves the claim for every finite number of vectors.

The empty combination gives $0\in\operatorname{span}(S)$. If $u$ and $w$ are finite linear combinations of vectors in $S$, concatenating their lists of terms expresses $u+w$ as another finite linear combination. For $\gamma\in F$, multiplying every coefficient in a finite combination for $u$ by $\gamma$ expresses $\gamma u$ in the span. The subspace test therefore applies.

If $W$ is any subspace containing $S$, then $W$ is closed under scalar multiplication and addition. It must therefore contain every linear combination of the vectors in $S$, so $\operatorname{span}(S)\subseteq W$. Hence the span is the smallest such subspace.