Definition: $(1)$ $B$ is called maximal linearly independent set if $B$ is independent and if $S\supset B$, then $S$ is dependent, or, equivalently, $\nexists S\supset B$ such that $S$ is independent. $(2)$ $B$ is called minimal spanning set if $\mathrm{span}(B)=V$ and if $S\subset B$, then $\mathrm{span}(S)\subset V$.
The following are equivalent:
$(1)$ $B$ is a maximal linearly independent set.
$(2)$ $B$ is minimal spanning set.
$(3)$ $B$ is basis of $V$.
Now there are lots of way to prove this theorem, like $(1)\Rightarrow (2)\Rightarrow (3) \Rightarrow (1)$, I think this is the least number of implication needed to prove this theorem(Note I’m not saying it is the most easiest way to do it) and $(1)\Leftrightarrow (2)$, $(1)\Leftrightarrow (3)$, $(2) \Leftrightarrow (3)$, this is maybe the most number of implication needed to prove this theorem. IMO, “easiest” way to prove this theorem, is to show $(1)\Leftrightarrow (3)$, $(2) \Leftrightarrow (3)$. You would agree proof of $(1)\Leftrightarrow (2)$ is essentially same as proof of $(1)\Leftrightarrow (3)$, $(2) \Leftrightarrow (3)$.
My attempt: $(1)\Rightarrow (2)$. Suppose $B$ is maximal independent set. Let $v\in V$. If $v\in B$, then $v\in B\subseteq \mathrm{span}(B)$. If $v\notin B$. Since $B\cup \{v\} \supset B$ and $B$ is maximal independent set, we have $B\cup \{v\}$ is dependent. Lemma: suppose $S_0\subseteq V$ is independent and $\beta\notin S_0$. Then $S_0\cup \{\beta \}$ is dependent$\iff$$\beta \in \mathrm{span}(S_0)$. Precisely, to prove $(\Rightarrow )$ implication we need $S_0$ is independent condition, to prove $(\Leftarrow )$ implication we need $\beta\notin S_0$ condition. By lemma, $v\in \mathrm{span}(B)$. Thus $V\subseteq \mathrm{span}(B)$. Hence $\mathrm{span}(B)=V$. Let $S\subset B$. Then $\exists u\in B\subseteq V$ such that $u\notin S$. We need to show $\mathrm{span}(S)\neq V$. Assume towards contradiction, $\mathrm{span}(S)= V$. Since $S\subset B$ and $B$ is independent, $S$ is independent. So $u\in V=\mathrm{span}(S)$. By lemma, $S\cup \{u\}$ is dependent. Since $S\cup \{u\} \subseteq B$ and $S\cup \{u\}$ is dependent, we have $B$ is dependent. Thus we reach contradiction. Hence $\mathrm{span}(S)\neq V$, i.e. $\mathrm{span}(S)\subset V$.
$(1)\Leftarrow (2)$. Suppose $B$ is minimal spanning set. We need to show $B$ is independent. Assume towards contradiction, $B$ is dependent, i.e. $\exists v\in B$ such that $v\in \mathrm{span}(B-\{v\})$. Claim: $\mathrm{span}(B-\{v\})=V$. Proof: let $x\in V=\mathrm{span}(B)$. Then $x=\sum_{i\in J_n}a_i \cdot u_i$, where $n\in \Bbb{N}$, $a_i\in F$, $u_i\in B$. If $v\notin \{u_i|i\in J_n\}$, then $x\in \mathrm{span}(B-\{v\})$. If $v\in \{u_i|i\in J_n\}$, then $v=u_j$, for some $j\in J_n$. So $x= \sum_{i\in J_n}a_i \cdot u_i$$= \sum_{i\in J_n-\{j\}}a_i \cdot u_i +a_j\cdot v$. Since $v\in \mathrm{span}(B-\{v\})$, we have $v= \sum_{i\in J_m}b_i \cdot w_i$, where $m\in \Bbb{N}$, $b_i\in F$, $w_i\in B-\{v\}$. So $x= \sum_{i\in J_n-\{j\}}a_i \cdot u_i + \sum_{i\in J_m}(a_j b_i) \cdot w_i$$\in \mathrm{span}(B-\{v\})$. Thus $V\subseteq \mathrm{span}(B-\{v\})$. Hence $V= \mathrm{span}(B-\{v\})$. Which contradicts the fact that $B$ is minimal spanning set. Our initial assumption must be wrong. Hence $B$ is independent. Let $S\supset B$. Then $\exists u\in S\subseteq V$ such that $u\notin B$. So $u\in V=\mathrm{span}(B)$. By lemma, $B\cup \{u\}$ is dependent. Since $B\cup \{u\}\subseteq S$ and $B\cup \{u\}$ is dependent, we have $S$ is dependent. Is my proof correct?
Note: The claim we proved in $(1) \Leftarrow (2)$ is quite standard result. We can generalize it, If $u\in \mathrm{span}(S-\{u\})$, then $\mathrm{span}(S)= \mathrm{span}(S-\{u\})$.