Am I sub-indexing this linear algebra proof correctly? Proposition
Let $ S \subset V \text{ and } w \in V$
$\textbf{span}(S \cup \{w\}) = \textbf{span}(S) \iff w \in \textbf{span}(S)$
Proof
$\impliedby$
Assume $w \in \textbf{span}(S)$.
We will check that if $v \in \textbf{span}(S \cup \{w\}) \text{ then } v \in \textbf{span}(S)$.
$v \in \textbf{span}(S \cup \{w\}) \implies \exists u_1,...,u_k \in S \text{ and } \exists a_1,...,a_k,b \in F: v = a_1u_1+...+a_ku_k+bw$
$w \in \textbf{span}(S) \implies w \text{ is a linear combination of vectors} \in S \implies$
$\exists u_1,...,u_k+u_{k+1},...,u_m \in S \text{ and } b_1,...,b_k,b_{k+1},...,b_m \in F:$
$w = b_1u_1+..+b_ku_k+b_{k+1}u_{k+1}+...+b_mu_m$
$v = (a_1+bb_1)u_1+...+(a_k+bb_k)u_k+(bb_{k+1})u_{k+1}+...+bb_mu_m$
Thus, $V \in \textbf{span}(S)$
$\implies$
Assume $\textbf{span}(S \cup \{w\}) = \textbf{span}(S) \implies w \in \textbf{span}(S \cup \{ w \}) \subset \textbf{span}(S)$
Q.E.D.
 A: Here it is an alternative approach to the case in which $S$ finite.
To begin with, let us reinforce the definition of a vector space spanned by a set. Given a finite dimensional vector space $V$ over the field $\textbf{F}$ and a set $S\subseteq V$, the vector space spanned by $S$ is the intersection of all subspaces from $V$ which contains $S$. That is the same as the set of all linear combinations of its elements. Having said that, we may proceed.
Let $S = \{s_{1},s_{2},\ldots,s_{n}\}$. If $\text{span}(S\cup\{w\}) = \text{span}(S)$, it means that any linear combination of the elements from the set $S\cup\{w\} = \{s_{1},s_{2},\ldots,s_{n},w\}$ also belongs to the set $\text{span}(S)$. In other words, we have that
\begin{align*}
& a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s_{n} + w = b_{1}s_{2} + b_{2}s_{2} + \ldots + b_{n}s_{n} \Longrightarrow\\\\
& w = (b_{1}-a_{1})s_{1} + (b_{2}-a_{2})s_{2} + \ldots + (b_{n}-a_{n})s_{n}
\end{align*}
from whence it results that $w\in\text{span}(S)$.
Conversely, the inclusion $\text{span}(S)\subseteq\text{span}(S\cup\{w\})$ is trivial. Indeed, if $s\in\text{span}(S)$, one has
\begin{align*}
s = c_{1}s_{1} + c_{2}s_{2} + \ldots + c_{n}s_{n} = c_{1}s_{1} + c_{2}s_{2} + \ldots + c_{n}s_{n} + 0w \Longrightarrow s\in\text{span}(S\cup\{w\})
\end{align*}
Let us now prove that $\text{span}(S\cup\{w\}) \subseteq \text{span}(S)$ based on the assumption $w\in\text{span}(S)$. To begin with, we should notice that
\begin{align*}
w = a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s_{n}
\end{align*}
Consequently, if $s\in\text{span}(S\cup\{w\})$, we have that
\begin{align*}
s & = b_{1}s_{1} + b_{2}s_{2} + \ldots + b_{n}s_{n} + b_{n+1}w\\\\
& = b_{1}s_{1} + b_{2}s_{2} + \ldots + b_{n}s_{n} + b_{n+1}(a_{1}s_{1} + a_{2}s_{2} + \ldots + a_{n}s_{n})\\\\
& = (b_{1} + b_{n+1}a_{1})s_{1} + (b_{2} + b_{n+1}a_{2})s_{2} + \ldots + (b_{n} + b_{n+1}a_{n})s_{n} \Longrightarrow s\in\text{span}(S)
\end{align*}
and the result holds.
A: All ok for me. You probably noticed that the proof essentially boils down to the fact that $A\subseteq span(B)\iff span(A)\subseteq span(B)$.
Here is an alternative:
\begin{align*}
span(S\cup\{w\})= span(S)\iff&span(S\cup\{w\})\subseteq span(S)\\
&\text{(because the reverse inclusion is always valid)}\\
\iff&S\cup\{w\}\subseteq span(S)\\
&\text{(by the fact above)}\\
\iff&\left\{w\right\}\subseteq span(S)\\
&\text{(because $S\subseteq span(S)$ always holds)}\\
\iff&w\in span(S)\\
&\text{(basic set theory)}
\end{align*}
