Ok...so I actually answered my own question with the help of Sheldon Axler (who commented). We know $B_1=\{u_1,u_2,w_1,...,w_n\}$ is linearly dependent because $u_2$ can be written as a linear combination of the vectors in $B$. This would like like so: $$u_2=c_0u_1+c_1w_1+...+c_nw_n$$ It follows that there are vectos $w_i\in B$ with $c_i\neq 0$ and at least one of these vectors is not $u_1$ because that would mean $u_2$ was a multiple of $u_1$ which Sheldon Axler pointed out is not possible. So if we fix a $w_j$ with $c_j\neq 0$ we can write: $$w_j=(-u_2+c_0u_1+c_1w_1+...+c_nw_n)*-c_j^{-1}$$ And now we can remove $w_j$ from $B_1$ and the resulting set will still span $V$.
The main thing to notice is the order of writing the list.
Your step 1 said we can remove $u_1$ is false. Note that in Linear Dependence Lemma, it said there exists a vector $v_j$ in the list so it belongs to span of all previous vectors $v_1, \ldots, v_{j-1}$. Notice all $v_1, \ldots, v_{j-1}$ are at the left of $v_j$ in the list $(v_1, \ldots, v_m)$.
Hence, if we write the list as
$$u_1,w_1, \ldots, w_n$$
in this exact order, we can't remove $u_1$ because all $w_i$ are at the right of $u_1$ in the list. Hence, even if $u_1$ is linear combination of $w_1, \ldots,w_n$, $u_1$ is still not the corresponding $v_j$ satisfying Linear Dependence Lemma.
Why can't we remove $u_i$?
All the steps give a general list $$u_1, \ldots, u_k,w_1, \ldots, w_l$$
that is linearly dependent. Linear Dependence Lemma said that in this this, there exists a vector $v$ so $v$ belongs to span of all previous vectors in the list. If this $v=u_i$ for some $i$ then $v \in \text{span}(u_1, \ldots, u_{i-1})$, which contradicts to condition that $(u_1, \ldots, u_k)$ is linearly independent. Thus, this $v$ must be different from any $u_i$. Thus, $v=w_i$ for some $i$. With this and condition (b) of the Linearly Dependent Lemma, we find that we can remove $v=w_i$. Thus, we can't remove $u_i$.
Best Answer
Here is a slight modification of the argument that might help you in understanding Axler’s.
Since the span of a list does not depend on the ordering, we can reorder lists in any way we like.
I’ll show that if we have replaced $k$ vectors in the spanning list with the first $k$ vectors in the linearly independent list and $k<m$, we’re able to replace one more vector.
The case in which $m=0$ is trivial, so we can assume $m>0$. Up to a reordering of the spanning list, we have a spanning list $$ (u_1,\dots,u_k,w_{k+1},\dots,w_n) $$ and therefore $u_{k+1}=a_1u_1+\dots+a_ku_k+a_{k+1}w_{k+1}+\dots+a_nw_n$ for some scalars $a_1,\dots,a_n$. Note that one among $a_{k+1},\dots,a_n$ is nonzero, otherwise we’d get a contradiction. Up to a further reordering, we may assume $a_{k+1}\ne0$ and therefore in the spanning list $$ (u_1,\dots,u_k,u_{k+1},w_{k+1},\dots,w_n) $$ we can remove $w_{k+1}$ to get the spanning list $$ (u_1,\dots,u_k,u_{k+1},w_{k+2},\dots,w_n) $$
Objection: how do we know that we still have a $w$-vector available? Let’s see. May we have $k<m$ and have exhausted the $w$-vectors? No. This would mean that $(u_1,\dots,u_k)$ is spanning and so $u_{k+1}$ would be a linear combination of $u_1,\dots,u_k$: contradiction.