Prove span of S is a subspace I would like to know if my proof is correct/reasonable.

I wish to prove the following: Let $V$ be a vector space over $F$. and $S$ is a subset of $V$. Prove $span(S)$ is a subspace of $V.$

I just want to know whether I am on the right track or I am completely off. Thanks

My solution:

Suppose $V$ is a vector space over some field $F$ that contains the vectors $v_{1}, v_{2}, .., v_{k}, .., v_{n}$. Take S to be the subset $\{v_{1}, v_{2}, .., v_{k}\}$

Then $span(S) = c_{1}v_{1}+c_{2}v_{2}+ .. +c_{k}v_{k}$ for some $c_{1}, c_{2},.., c_{k} \in F$. We want to show that this is a subspace of V.

Clearly, ${\bf0} \in span(S)$ when $c_{1},..,c_{k} = 0$.

Suppose ${\bf x}, {\bf y}\in span(S)$ where ${\bf x} = c_{1}v_{1}+c_{2}v_{2}+..+c_{k}v_{k}$ and ${\bf y} = d_{1}v_{1}+d_{2}v_{2}+..+d_{k}v_{k}$ for scalars $c_{1},..,c_{k}, d_{1},..d_{k} \in F$. Also let $\lambda \in F$.

Consider ${\bf x} + \lambda {\bf y} = c_{1}v_{1}+c_{2}v_{2}+..+c_{k}v_{k} + \lambda (d_{1}v_{1}+d_{2}v_{2}+..+d_{k}v_{k})$

$=(c_{1}+\lambda d_{1})v_{1}+(c_{2}+\lambda d_{2})v_{2}+..+(c_{k}+\lambda d_{k})v_{k}$. 

Since $(c_{k}+\lambda d_{k}) \in F$, ${\bf x} + \lambda {\bf y} \in span(S)$. Hence $span(S)$ is a subspace of $V.$
Hints: 

The span $[S]$ by definition is the intersection of all sub - spaces of $V$ that contain $S$. Use this to prove all the axioms if you must. The identity exists in every subspace that contain $S$ since all of them are subspaces and hence so will the intersection. The Associativity law for addition holds since every element in $[S]$ is in $V$. The additive inverse also exists by the argument that any element in $[S]$ is also in every subspace that contains $S$ and the inverse exists in each of these subspaces and hence in the intersection. Argue down a similar path. 

This is a lot more easier if you've been taught a Theorem to characterise subspaces.
Hints: 

The span $[S]$ by definition is the intersection of all sub - spaces of $V$ that contain $S$. Use this to prove all the axioms if you must. The identity exists in every subspace that contain $S$ since all of them are subspaces and hence so will the intersection. The Associativity law for addition holds since every element in $[S]$ is in $V$. The additive inverse also exists by the argument that any element in $[S]$ is also in every subspace that contains $S$ and the inverse exists in each of these subspaces and hence in the intersection. Argue down a similar path. 

This is a lot more easier if you've been taught a Theorem to characterise subspaces.