Linearly independent Let $S$ be a linearly independent subset of a vector space $V$, and let $v$ be a vector in $V$ that is not in $S$. Then $S\cup \{v\}$ is linearly dependent if and only if $v\in span\{S\}$.  
proof)
If $S\cup \{v\}$ is linearly dependent then there are vectors $u_1,u_2,\dots,u_n$ in $S\cup \{v\}$ such that $a_1u_1+a_2u_2+\dots+a_nu_n=0$ for some nonzero scalars $a_1,a_2,\dots,a_n$.
Because $S$ is linearly independent, one of the $u_i$'s, say $u_1$ equals $v$.  
[ADDITION]
The last part of the proof is this: Because S is linearly independent, one of the $u_i$'s, say $u_1$ equals $v$. Thus $a_1v+a_2u_2+\dots+a_nu_n=0$ and so $v$ can be written as a linear combination of $u_2,\dots,u_n$ which are in $S$. By definition of span, we have $v\in span(S)$. 
I can't understand the last sentence. I think since $S$ is linearly independent and $S\cup \{v\}$ is linearly dependent so consequently $v$ can be written as the linear combination of $u_1,u_2,\dots,u_n$. But it has any relation to that sentence?  
(+) I also want to ask a simple question here.
Any subset of a vector space that contains the zero vector is linearly dependent, because $0=1*0$. But that shows it holds when there is only one vector, zero vector, and the coefficient $a_1=1$.
Then it still holds when there are other nonzero vectors in a vector space?  

Your intuition is good but there is a problem with stating that $v$ is in the span of $S$ directly.  Mainly  if we have some linear combination:
$$(\ast) \qquad a_1s_1 + \cdots a_ns_n + a_{n+1}v = 0$$
where not all the $a_i$ are zero, we want to show that $a_{n+1}$ is not zero so that we can subtract that term over and divide by $-a_{n+1}$ That is,
$$a_1s_1 + \cdots a_ns_n  = -a_{n+1}v$$
and so to finish the proof we only need to show that $-a_{n+1}$ is not zero so that we can divide by it.  If $a_{n+1}$ were zero then $(\ast)$ would have a nontrivial linear combination equal to 0 which contradicts the fact that $S$ is linearly independent. So $a_{n+1}$ is not zero and we can divide by it so that $v$ is in the span of $S.$ 
This is essential the step that your proof above is using.  If you have some combination that is zero then one of the terms must be $v$ an it must have a nonzero coefficient by the linearly independence of $S.$
