Correctly negating "there exists a subset of $S$ that is a basis for $V$" I would like to prove the following by contradiction: 
"Let $V$ be a vector space having dimension $n$, and let $S$ be a subset of $V$ that generates $V$. Prove that there is a subset of $S$ that is a basis for $V$."
My textbook's definition of a basis: A basis $\beta$ for a vector space $V$ is a linearly independent subset of $V$ that generates $V$.
I would like to negate that there is a subset of $S$ that is a basis for $V$:
There does not exist a subset of $S$ that is a basis for $V$. 
$\implies$ For all $A \subset S$, $A$ is not a basis. 
$\implies$For all $A \subset S$, $A$ is not linearly independent OR $A$ does not generate $V$.
$S$ is a subset of $S$ that generates $V$.
$\implies$ For all $A \subset S$, $A$ is not linearly independent.
$\implies$ There does exist a linearly independent subset of $S$.
$\implies$ Every subset of $S$ is linearly dependent.
However, my professor says that the correct negation is: "Any subset of S with $n$ vectors is linearly dependent."
Using just the definitions, I'm not sure how that is justified. I'm wondering where my chain of reasoning is wrong.
 A: Your professor would have meant " Any subset of $S$ with more than $n$ vectors is linearly dependent " but I do not see how you cold use this to prove your statement. The negation for the claim "There is a subset of $S$ which is a basis" is as you said " There does not exist a subset of $S$ that is a basis for $V$ " which is not necessarily useful. A better, equivalent way to say the same would be "No subset of $S$ is a basis for it", the meaning of which you seem to have unearthed. 

But this is not how I would go about proving this. First pick the
  smallest subset of $S$ which would generate $V$. Then argue that it has to be linearly independent. Then this smallest set will be a basis.

A: It is common knowledge that no subset of a vector space of dimension $n$ can be generated by fewer than $n$ vectors (while it can be generated with $n$ or more vectors).  
Some subsets of $S$ with fewer than $n$ vectors may very well be linearly independent, (hence failing to generate $V$), and no contradiction results. (E.g.) A subset of one vector is linearly independent, though it doesn't generate all of $V$ if $n \gt 1$.
Since you are using the hypothesis: "$A$ is a subset of $S$ that generates $V$," you are implicitly claiming also that $S$ has dimension equal to or greater than $n$. 
I'll use your proof as a framework, but annotate and correct the line of reasoning:

There does not exist a subset of $S$ that is a basis for $V$. 
$\implies$ For all $A \subset S$, $A$ is not a basis. 
$\implies$For all $A \subset S$, $A$ is not linearly independent OR $A$ does not generate $V$.
$A$ is a subset of $S$ that generates $V$,  
$\qquad \implies |A| \geq n$.
$\implies$ For all $A \subset S$, $|A| \geq n \implies A$ is not linearly independent. (We already know that if $|A|\gt n \implies A$ is not linearly independent.)
$\implies$ There does not exist a subset with $n$ vectors which is a linearly independent subset of $S$.
$\implies$ Every subset of $S$ with $n$ vectors is linearly dependent.
A: You can use your professor's statement this way:
First prove a usefull lemma:
$\textbf{Lemma:}$ Let $V$ be a $n$-dimensional vector space. Every $n$-elements linear independent subset of $V$ is a basis.
$\textbf{Proof of lemma:}$
Suppose not. There exists a linear independent subset $A$ of $V$ that $|A|=n$ and isn't a basis.
According your definision of basis $A$ doesn't generate $V$  or is linear dependent. But you know that $A$ is linear independent, so it doesn't generate $V$. So there exists vector $v \not\in \text{span}(A)$ and $v \in V$. 
Now because $v \not\in \text{span}(A)$, then $B=A \cup \{v\}$ is linear independent (it's possible to prove if it isn't obvious).$B$ is set of $n+1$ linear independent vectors, so
$$\dim \left(\text{span}(B)\right) = n+1$$
But you know that $B \subset V$, so $\text{span}(B) \subset \text{span}(V)=V$, so:
$$\dim \text{span}(B) \leq \dim V$$
But $\dim \text{span}(B)=n+1$ and $\dim V=n$. $\square$
Now you can use this lemma to prove your statement.
$\textbf{Prove of statement:}$
Suppose that there does not exist a subset of $S$ that is a basis of $V$. Then by your definition every subset fo $S$ is linear dependent of doesn't generate $V$. Let $A$ be a linear independent subset of $V$, that $x=|A|$ is the largest possible. Now because $\dim V=n$, then $x \leq n$. If $x=n$ then $A$ is a basis by lemma. So $x<n$. 
But $\dim \text{span}S=n$, so there exists vector $v \in S$ and $v \not\in \text{span}(A)$. So $B=A \cup \{v\}$ is linear independent (like in lemma's proof). But $|B|>x$. It's contradiction, because $x$ is the largest possible.
