Why do we know that a set S of sentences implies the sentence A if the union of S and the negation of A is inconsistent? As in the title: Why do we know that a set $S$ of sentences implies the sentence $A$ if the set $S \cup \lnot A$ is inconsistent?
I "know" it's because if $S \cup \lnot A$ is inconsistent, then $S \cup A$ must be consistent. But why must $S \cup A$ be consistent if $S \cup \lnot A$ is inconsistent? Do we assume that $S$ is consistent, and notice that if we add $\lnot A$ to it, the union is inconsistent, and from there deduce that if instead of $\lnot A$ we would've added $A$, the set would've been consistent? If so, how can we deduce this? What makes it impossible for $S \cup \lnot A$ and $S \cup A$ both to be inconsistent or consistent?
I don't know if it matters, but I'm talking about basic propositional logic.
 A: 
I "know" it's because if $S\cup\{\neg A\}$ is inconsistent, then $S\cup\{A\}$ must be consistent.

No. That's not why it's the case, and indeed that statement is false (e.g. take $S$ itself to be something inconsistent). It is true if we assume $S$ is consistent, but we don't need to do this.
The point is that proof by contradiction is built into classical logic: in $S$, we can argue roughly as follows: 

"Suppose $\neg A$. Then (since we're assuming that $S\cup\{\neg A\}$ is inconsistent) we can prove a contradiction. So our hypothesis $\neg A$ was false, hence $A$ is true."

Note that a consequence of proof by contradiction is that an inconsistent theory proves every sentence.
A: To say that $S \cup \{ \lnot A \}$ is inconsistent is to say that it is unsatisfiable, i.e. there is no valuation $v$ such that $v(\sigma)=$ t for every $\sigma \in S$ and $v(\lnot A)=$ t.
Consider now a valutaion $v$ such that $v(\sigma)=$ t for every $\sigma \in S$.
By the above we must have $v(\lnot A)=$ f, and thus $v(A)=$ t.
This implies that:

$S \vDash A$.

