Proving a lemma - show the span of a union of subsets is still in the span This is part of proving a larger theorem but I suspect my prof has a typo in here (I emailed him about it to be sure)
The lemma is written as follows: 
Let $V$ be a vector space. Let {$z, x_1, x_2, ..., x_n$} be a subset of $V$. Show that if $z \in\ span(${$x_1, x_2,..., x_n$}$)$, then $span(${$z, x_1, x_2,..., x_r$}$)=span(${$ x_1, x_2,..., x_r$})
I feel like this should be really simple and I saw a proof that you can take out a vector from a subset and not change the span, but I am unsure of the reverse -- assuming that is what this lemma is about. (To me, the "if" implies something besides just unifying the two sets should follow). 
Anyhow, the proof should, I think, start with that we can modify a subset (let's call it S) without affecting the span if we start like this: 
$\exists x \in S$ such that $ x \in span(S-${$x$})
then you build up a linearly independent subset somehow. The proof that you can take vectors out of the subset says that since $x\in span(${$x$}$)\ \exists\ \lambda_1, \lambda_2,..., \lambda_n \ \in\ K$ such that $x=\sum_{i=1}^n\lambda_i x_i $
since we know $ span(S) \supset span(S-${$x$}) we just need to show $ span(S) \subseteq span(S-${$x$})
But honestly I am not sure I understand what's happening here well enough to prove the above lemma. (This class moves fast enough that we're essentially memorizing proofs rather than re-deriving them I guess). 
I am really starting to hate linear algebra. :-( 
(Edited to fix U symbol and make it a "is a member of" symbol)

We want to show $\operatorname{span}\{z, x_1, \dots, x_n\} = \operatorname{span}\{x_1, \dots, x_n\}$.   In general, to show $X = Y$ where $X, Y$ are sets, we want to show that $X \subseteq Y$ and $Y \subseteq X$.
So suppose $v \in \operatorname{span}\{x_1, \dots, x_n\}$.  Then, we can find scalars $c_1, \dots, c_n$ such that $$v = c_1x_1  + \dots + c_nx_n$$ so clearly, $v   \in \operatorname{span}\{z, x_1, \dots, x_n\}$.  This proves $$\operatorname{span}\{x_1, \dots, x_n\} \subseteq \operatorname{span}\{z, x_1, \dots, x_n\}$$
Now let $v \in \operatorname{span}\{z, x_1, \dots, x_n\}$.  Again, by definition, there are scalars $c_1, \dots, c_{n+1}$ such that $v = c_1x_1 + \dots + c_nx_n + c_{n+1}z$. But hold on, $z \in \operatorname{span}\{x_1, \dots, x_n\}$, right? This means there are scalars $a_1, \dots, a_n$ such that $z = a_1x_1 + \dots + a_nx_n$.  Hence,
$$v = c_1x_1 + \dots + c_nx_n + c_{n+1}(a_1x_1 + \dots + a_nx_n)$$
$$= (c_1 + c_{n+1}a_1)x_1 + \dots + (c_n+c_{n+1}a_n)x_n$$
and so we conclude that $v \in \operatorname{span}\{x_1, \dots, x_n\}$. Therefore, 
$$\operatorname{span}\{z, x_1, \dots, x_n\} = \operatorname{span}\{x_1, \dots x_n\}$$
