$w$ is in the span Is the following theorem correct? I will especially appreciate any comments on how to improve the writing of the proof.
NOTE: In the proof below $\mathbf{2.21-(a)}$ is the result that given a linearly dependent list $v_1,v_2,...,v_m$
$$\exists j\in\{1,2,...,m\}(v_j\in span(v_1,v_2,...,v_{j-1}))$$
Theorem. Given that $v_1,v_2,...,v_m$ is a linearly independent in $V$ and $w\in V$. Show that $v_1,v_2,...,v_m,w$ is linearly independent if and only if $$w\not\in span(v_1,v_2,...,v_m).$$
Proof. $(\Rightarrow).$ Assume for purpose of contradiction that $v_1,v_2,...,v_m,w$ is a linearly independent list in $V$, $w\in V$ and $w\in span(v_1,v_2,...,v_m)$. It follows that 
$$w=\sum_{i=1}^{m}a_iv_i,\ \forall i\in\{1,2,3,..,m\}(a_i\in\mathbf{F})\tag{1}$$
but this implies that $v_1,v_2,...,v_m,w$ is not linearly independent resulting in a contradiction.
$(\Leftarrow).$ We prove the contrapositive, assume that $u_1,u_2,...,u_{m},u_{m+1}=v_1,v_2,...,v_m,w$ is a linearly dependent list, then by $\mathbf{2.21-(a)}$ it follows that for some $j\in I=\{1,2,...,m,m+1\}$, $u_j\in span(u_1,u_2,...,u_{j-1})$. Consider now the following Lemma.
Lemma $j=m+1$
Proof. Assume on the contrary that $j\neq m+1$ it then follows that $j=k$ for some $k\in I\backslash\{m+1\}$ consequently $u_j=v_k\in span(v_1,v_2,...,v_{k-1})$, moreover $v_1,v_2,...,v_m$ is a linearly independent on $V$ consequently $v_1,v_2,...,v_k$ is linearly independent on $V$ but then $v_k\not\in span(v_1,v_2,...,v_{k-1})$ resulting in a contradiction.
$\square$
From the above lemma it follows that $u_j=u_{m+1}=w$ implying that $w\in span(v_1,v_2,...,v_m)$.
$\blacksquare$
 A: The proof is correct as written, but you can make it an $\epsilon$ clearer.
For instance, when you write $w\in\operatorname{span}(v_1,\dots,v_m)$, you note correctly that there are scalars $a_i$ such that
$$
w = \sum a_iv_i.
$$
Now note that this implies directly that
$$
w - \sum a_iv_i = 0,
$$
which is a nontrivial dependence relation (note the invisible $1$ in front of $w$), contradicting what you assumed that the list $(v_1,\dots,v_m,w)$ is independent.
For your other direction, you are given the key lemma that in any ordered dependent list $(v_1,\dots,v_m)$ there is an index $i$ such that $v_i$ is in the span of the preceding vectors. As a matter of style, you don't need this extra lemma that $j = m+1$; just go ahead and show it with your key lemma as follows:
If $(v_1,\dots,v_m,w)$ is a dependent list, then by the key lemma, there is a vector in the list which is in the span of the preceding vectors. As $(v_1,\dots,v_m)$ is known to be independent, we deduce that $w$ is in the span of $(v_1,\dots,v_m)$.
This is easier to follow because it doesn't weigh down your readers' eyes with superfluous information.
