trying to understand a proof of menger's theorem 
I'm trying to understand the short proof on Menger's Theorem from Wikipedia. I have several fairly (basic) questions that are preventing me from understanding the proof well, shown below.

This is probably a basic question, but in this proof, are they assuming a minimum AB-separator for G has size k?
I don’t understand how one can find an S to be an AB-separator of size less than k so that every AB-path in G has a vertex of S or the edge e (why does this S necessarily exist).
Also, why is it true that S together with either endpoint of e would be a better AB-separator of G? It clearly is larger, but why exactly would it be an AB-separator?
Why is S alone too small to be an AB-separator of G?
Why is it that because of the size of $C_1$, the endpoints of the paths in it must be exactly $S_1$?
Why is it true that every path in $C_1$ is internally disjoint from every path in $C_2$ because $S_1$ disconnects G? I was wondering if one could justify this by showing that if there were two paths $P_1$ in $C_1$ and $P_2$ in $C_2$ that weren’t internally disjoint, one would get a contradiction?

Even helping me to resolve one of these questions would be useful.

 A: If you go back and re-read the proof, you’ll see that an $AB$-separator of $G-e$ of cardinality less than $k$ doesn’t necessarily exist: as it says in the second paragraph of the proof, there might be one of cardinality $k$, in which case the induction hypothesis gives us an $AB$-connector of cardinality $k$ in $G-e$ and hence the desired one in $G$. It’s the rest of the proof that deals with the case in which there is no $AB$-separator of cardinality $k$ in $G-e$.
In that case $G-e$ must have an $AB$-separator $S$ of cardinality less than $k$. By definition of $AB$-separator every $AB$-path in $G-e$ goes through $S$, i.e., has a vertex in $S$. $S$ is too small to be an $AB$-separator in $G$, because by hypothesis the minimal cardinality of an $AB$-separator in $G$ is $k$. Thus, there must be $AB$-paths in $G$ that don’t go through $k$. The only edge that we removed was $e$, so those paths must all use the edge $e$: that’s the only way that they could exist in $G$ but not in $G-e$. And if they use $e$, they must pass through $v_1$ and $v_2$, the endpoints of $e$. That means that every $AB$-path in $G$ goes through $S\cup\{v_1\}$, and every $AB$-path in $G$ goes through $S\cup\{v_2\}$, i.e., that both $S\cup\{v_1\}$ and $S\cup\{v_2\}$ are $AB$-separators in $G$.
If $|S|<k-1$, then $|S\cup\{v_1\}|<k$, and $S\cup\{v_1\}$ would then be an $AB$-separator in $G$ of cardinality less than $k$. (The same holds for $S\cup\{v_2\}$.) By hypothesis, however, no such $AB$-separator exists. (The word better in the proof was poorly chosen: what is meant is smaller.) Thus, $|S|$ must be at least $k-1$. We were already assuming that $|S|\le k-1$, so in fact we must have $|S|=k-1$.
$C_1$ is an $AS_1$-connector of cardinality $k$: it consists of $k$ vertex-disjoint $AS_1$ paths. Since these $k$ paths have no vertices in common, their endpoints in $S_1$ must be $k$ different vertices. But $S_1$ contains only $k$ vertices, so each of these vertices must be the endpoint of one of the paths making up $C_1$.
Let $P_1$ be one of the paths making up $C_1$ and $P_2$ one of the paths making up $C_2$. Let $v$ be any internal vertex of $P_1$, and suppose that $v$ is also an internal vertex of $P_2$. Let $a$ be the endpoint of $P_1$ in $A$ and $b$ the endpoint of $P_2$ in $B$. Then the part of $P_1$ from $a$ to $v$ together with the part of $P_2$ from $v$ to $b$ would be an $AB$-path that did not go through $S_1$, which is impossible, since $S_1$ is an $AB$-separator. Thus, the paths making up $C_1$ must be internally vertex-disjoint from the paths making up $C_2$.
(The schematic diagram that accompanies the proof is actually quite helpful.)
