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
Since $\{w_1,\ldots,w_n\}$ spans $V$, and $u_1\in V$, there exist $a_i$ such that $u_1=a_1w_1+\cdots+a_nw_n$. So $(-1)u_1+a_1w_1+\cdots+a_nw_n=0$ and therefore the adjoined set is linearly dependent.