I've little experience with Zorn's lemma, and my class hasn't covered the material yet. However, the teacher has stated that the claim holds for infinite dimensional vector spaces, and he implies he wants us to use this fact on our homework. I was not satisfied with this. I felt we should justify this claim before using it.
$\textbf{Proof:}$
Let $V$ be a vector space over a field $F$. Let $S\in V$ be a linearly independent set. Consider the set
$$\mathcal{S}=\{S'\subseteq V|S\subseteq S'\text{ and }S'\text{ is linearly independent.}\}.$$
This set is partially ordered by subset inclusion. Consider a chain in $\mathcal{S}$
$$S_1\subset S_2\subset…\subset S_i\subset…$$
which may or may not terminate. It follows that $\{S_i\}$ is bounded above by $\bigcup S_i$. We will show that $\bigcup S_i$ is a linearly independent set. If $\bigcup S_i$ were linearly dependent, then there exists $v_1,…,v_n\in\bigcup S_i$ such that $a_1v_1+…+a_nv_n=0$ has a non-trivial solution, that is $v_1,…,v_n$ are linearly dependent. Note that since $v_j\in \bigcup S_i$ it follows that there $S_{i_j}$ such that $v_j\in S_{i_j}$. Then $v_1,…,v_n\in S_m$ where $m=\max(i_1,…,i_n)$. This contradicts that $S_m$ is linearly independent, hence $\bigcup S_i$ is linearly independent, and belongs to $\mathcal{S}$. This shows that every totally ordered chain in $\mathcal{S}$ has an upper bound in $\mathcal{S}$, therefore by Zorn's Lemma $\mathcal{S}$ has a maximal element.
Let $S^m\in\mathcal{S}$ be maximal. Clearly $S\subseteq S^m$. If we can show $S^m$ is a spanning set, then since it is linearly independent ($S^m\in\mathcal{S}$) it will be a basis containing $S$ completing the proof.
Suppose that $v\in V$ such that $v\not\in\text{Span}(S^m)$. This happens if and only there exists no set of vectors $v_1,…,v_n\in S^m$ and scalars $a_1,…,a_n,a\in F$ not all $0$ such that
$$a_1v_1+…+a_nv_n=-av.$$
This implies that for all $v_1,…,v_n\in S^m$ the equation
$$a_1v_1+…+a_nv_n+av=0$$
has only the trivial solution, which happens if and only if $S^m\cup\{v\}$ is a linearly independent set. This is impossible, for $S^m$ is maximal, therefore by contradiction $S^m$ spans $V$.$\blacksquare$
Best Answer
The proof is good, apart from some minor details.
A chain in $\mathcal{S}$ is not necessarily of the form $S_1\subset S_2\subset\dotsb$. It is a subset $\mathcal{T}$ of $\mathcal{S}$ such that, for all $T_1,T_2\in\mathcal{T}$ either $T_1\subset T_2$ or $T_2\subset T_1$ ($\subset$ denoting nonstrict inclusion).
Two lemmas will be used.
Proof. Essentially, this is the definition. □
Proof. if a finite subset of $S$ is not linearly independent it must include $v$; so it is of the form $\{v_1,\dots,v_n,v\}$ and there are $a_1,\dots,a_n,a$ not all zero with $$ a_1v_1+\dots+a_nv_n+av=0 $$ Now $a\ne0$, because $S$ is linearly independent, so $$ v=(-a^{-1})a_1v_1+\dots+(-a^{-1})a_nv_n $$ contradicting that $v\notin\operatorname{span}(S)$. □
Consider a chain $\mathcal{T}$ and let $T=\bigcup\mathcal{T}$. Then $T$ is linearly independent. Indeed, if $v_1,\dots,v_n\in T$, there is a member $T_0$ of $\mathcal{T}$ such that $v_1,\dots,v_n\in T_0$ (easy induction, fill in the details). In particular $\{v_1,\dots,v_n\}$ is linearly independent. Lemma 1 allows to conclude that $T$ is linearly independent.
Therefore, Zorn's lemma guarantees the existence of a maximal element $\bar{T}\in\mathcal{S}$.
The fact that $\bar{T}$ is a spanning set follows Lemma 2.