I tried to construct proof of the theorem without actually looking at the proof. I would like to get verification and feedback for it.
Theorem 3
The subspace spanned by a non-empty subset $S$ of a vector space $V$ is the set of all linear combinations of vectors in $S$.
Theorem 3 proof
Let $U$ be the subspace spanned by a non-empty subset $S$ of a vector space $V$. So $U=\{\cap W |S\in W\subset V \}$.
We need to show that $U$ is the set o fall linear combinations of vectors in $S$. i.e
$$U=\{\sum_i\ c_i \alpha_i |\alpha_i \in S\text{ and } c_i \text{ belongs to the field}\}.$$
Proof
"$\Rightarrow$" Since $U$ is a subspace, all its elements are of the form
$$c\alpha+\beta \qquad \text{where $\alpha,\beta\in S,\; c\in F$}.$$
Since $U$ is the intersection of all subsets containing $S$, $U$ is the minimal subspace containing $S$.
By axiom, the sum of any vectors of $U\supset S$ is again in $U$. So $\sum_i\ c_i \alpha_i$ is in $U$ for $c=1$.
Also, for any $c\in F \text{ and } \alpha \in S, \quad c\alpha \in U$ by axiom. Hence by addition axiom for any $c\in F \text{ and } \alpha \in S$, $\sum_i\ c_i \alpha_i$ is also in $U$.
"$\Leftarrow$" Let $U$ be the set of all linear combinations of vectors in $S\subseteq V$.
We have to show that $U$ is the subspace spanned by a non-empty subset $S$ of a vector space $V$ i.e
$$U=\{\cap W |S\in W\subset V \}.$$
Since $U$ is the subspace and the set of all linear combinations of $S$, $S \subseteq U$. Because $1\alpha\in U$ for any $\alpha \in S$. Hence $U$ is intersection of sets containing $S$.
Definition
Let $S$ be a set of vectors in a vector space $V$. The subspace spanned by $S$ is defined to be the intersection $W$ of all subspaces of $V$ which contain $S$.