If $S \subseteq V$ is linearly independent and $w \notin \operatorname{span}(S)$ then $S \cup \{w\}$ is linearly independent and $\operatorname{span}(S) \subsetneq \operatorname{span}(S \cup \{w\})$; and if $w \in \operatorname{span}(S)$ then $\operatorname{span}(S \cup \{w\}) = \operatorname{span}(S)$

Claim 1. From $S \subseteq S \cup \{w\}$ we get $\operatorname{span}(S) \subseteq \operatorname{span}(S \cup \{w\})$. Conversely, assume $w \in \operatorname{span}(S)$; since also $S \subseteq \operatorname{span}(S)$, the set $S \cup \{w\}$ is contained in $\operatorname{span}(S)$, which is a linear subspace of $V$, so minimality gives $\operatorname{span}(S \cup \{w\}) \subseteq \operatorname{span}(S)$. The two inclusions give the claim.

The two easy parts of claim 2. Assume $w \notin \operatorname{span}(S)$. Then $w \notin S$, because $S \subseteq \operatorname{span}(S)$. Also $\operatorname{span}(S) \subseteq \operatorname{span}(S \cup \{w\})$ by monotonicity, and $w$ lies in the larger set and not in the smaller, so the inclusion is strict.

Now assume in addition that $S$ is independent, and let $v : n \to S \cup \{w\}$ be an injective finite list with $\lambda : n \to F$ and $\sum_{i<n}\lambda_i v_i = 0_V$. If $w$ is not a value of $v$, then $v$ is an injective finite list into $S$, so independence of $S$ gives $\lambda_i = 0_F$ for every $i < n$ and there is nothing more to prove.

In the remaining case $w = v_k$ for exactly one $k < n$, since $v$ is injective. Then $n \ne 0$, say $n = \sigma(n')$, and $y := v \circ \delta_k$ is an injective finite list $n' \to S$: it is injective as a composite of injections, and its values are the $v_j$ with $j \ne k$, each of which lies in $S \cup \{w\}$ and differs from $v_k = w$. Moreover, for every $\mu : n \to F$ with $\mu_k = 0_F$ the list $i \mapsto \mu_i v_i$ has the value $0_F w = 0_V$ at $k$, so deleting that index gives $\sum_{i<n}\mu_i v_i = \sum_{i<n'}\mu_{\delta_k(i)} y_i$.

In that case the coefficient of $w$ vanishes. Suppose $\lambda_k \ne 0_F$. Applying (F3) at $k$ to the list $i \mapsto \lambda_i v_i$ gives $0_V = \lambda_k w + R$, where $R = \sum_{i<n}\lambda'_i v_i$ with $\lambda'_k := 0_F$ and $\lambda'_i := \lambda_i$ for $i \ne k$, using $0_F w = 0_V$ to identify the deleted entry. By step 1.4 applied to $\lambda'$, $R = \sum_{i<n'}\lambda'_{\delta_k(i)} y_i$, which is a linear combination of elements of $S$ and therefore lies in $\operatorname{span}(S)$. Then $\lambda_k w = -R$ lies in $\operatorname{span}(S)$, that set being a linear subspace, and hence so does $w = 1_F w = (\lambda_k^{-1}\lambda_k)w = \lambda_k^{-1}(\lambda_k w)$, contradicting $w \notin \operatorname{span}(S)$. So $\lambda_k = 0_F$.

The remaining coefficients vanish too. Since $\lambda_k = 0_F$ by step 2.1, step 1.4 applied to $\lambda$ itself gives $0_V = \sum_{i<n}\lambda_i v_i = \sum_{i<n'}\lambda_{\delta_k(i)} y_i$; the list $y$ is an injective finite list into the independent set $S$, hence independent, so $\lambda_{\delta_k(i)} = 0_F$ for every $i < n'$. As $\delta_k$ has image $n \setminus \{k\}$, this says $\lambda_j = 0_F$ for every $j \ne k$, and with $\lambda_k = 0_F$ every coefficient vanishes.

Steps 1.3 and 3.1 show that every injective finite list into $S \cup \{w\}$ is independent, so $S \cup \{w\}$ is linearly independent; with step 1.2 this is claim 2, and step 1.1 is claim 1.