This is a part of the proof of
LEMMA 3.1. Each cut-of-the-phase is a minimum s-t-cut in the current graph,
where $s$ and $t$ are the two vertices added last in the phase.
$A_{v}$ the set of all vertices added before $v$, so $A_{v} \cup {v}$ the set of all vertices added before $v$ including $v$.
$C_{v}$ is just the cut of $A_{v} \cup {v}$ on all vertices of $G$. Obviously all $C_{v}$'s are different because they are cuts for different sets of vertices, but all these vertices are vertices of initial graph $G$ (with the same edges as edges of graph $G$, but of course $C_{v}$ has less edges that $G$, all vertices of $C_{v}$ are the vertices of $G$, but $G$ may have more vertices). So think of $C_{v}$ as a subgraph of $G$, there is no something new in $C_{v}$ that $G$ doesn't have.
The more official definition from Glossary of graph theory:
A subgraph $H$ of a graph $G$ is said to be induced if, for any pair of vertices $x$ and $y$ of $H$, $xy$ is an edge of $H$ if and only if $xy$ is an edge of $G$. In other words, $H$ is an induced subgraph of $G$ if it has exactly the edges that appear in $G$ over the same vertex set. If the vertex set of $H$ is the subset $S$ of $V(G)$, then $H$ can be written as $G[S]$ and is said to be induced by $S$.
Addendum:
Let's be more specific about the algorithm. I think wlog we can consider $C_{v}$ as $C$, it doesn't contradict the definition of induced graph and works great to show what the algorithm does. According to the algorithm we approach the last vertex by taking the largest in terms of weight edges. Therefore as the cut we have the lightest edges so obviously it's the minimum s-t in comparison to any arbitrarily taken s-t cut.
Please take a look at the following description Simple min cut algorithm. It provides with very good example.
These notes contain a proof. The digraph used in this proof is a little more complicated than the one that you have in mind: each point of the partial order corresponds to two vertices of the digraph.
Let $\langle P,\preceq\rangle$ be the partially ordered set; I’ll write $p\prec q$ to indicate that $p\preceq q$ and $p\ne q$. For each $p\in P$ the digraph $D$ will have two vertices, $p^-$ and $p^+$; in addition, it will have a source vertex $s$ and a sink vertex $t$. For each $p\in P$, $D$ has edges $\langle s,p^-\rangle$ and $\langle p^+,t\rangle$; in addition, for each$p,q\in P$ with $p\prec q$ it has $\langle p^-,q^+\rangle$. Each edge has capacity $1$.
Let $f$ be a maximal flow in $D$, and let $\langle S,T\rangle$ be a minimal cut constructed by the Ford-Fulkerson algorithm. Note that every path from $s$ to $t$ has the form $s\to p^-\to q^+\to t$ for some $p,q\in P$ with $p\prec q$; thus $|f|$ is the number of $p\in P$ such that $f(s,p^-)=1$ there is a $q\in P$ such that $f(p^-,q^+)=1$. Let $C=\{p\in P:f(s,p^-)=0\}$; for future reference note that $|C|=|P|-|f|$.
If $p\in P\setminus C$, then $f(s,p^-)=1$, and there must therefore be a unique ‘successor’ $\sigma(p)\in P$ such that $f(p^-,\sigma(p)^+)=1$, and hence $p\prec\sigma(p)$. Thus, for each $p\in P$ there is a well-defined chain $$p=p_0\prec p_1\prec\ldots\prec p_n\in C$$ in $P$ such that $p_{k+1}=\sigma(p_k)$ for $k=0,\dots,n-1$. ($P$ is finite, so each chain must terminate, and the elements of $C$ are the only elements without successors.) For $p\in P\setminus C$ let $\xi(p)$ be the unique element of $C$ at which the chain from $p$ terminates, and let $\xi(p)=p$ for $p\in C$; then $$\Big\{\{p\in P:\xi(p)=q\}:q\in C\Big\}$$ partitions $P$ into $|C|$ chains. To complete the proof we must show that $P$ also has an antichain of cardinality $|C|$.
The desired antichain is $A=\{p\in P:p^-\in S\text{ and }p^+\in T\}$. Proving that it’s an antichain isn’t too hard. Suppose that $p,q\in A$ with $p\ne q$. Then $p^-\in S$ and $q^+\in T$, so $f(p^-,q^+)\ne0$: either $\langle p^-,q^+\rangle$ isn’t an edge of $D$ at all, or $f(p^-,q^+)=1$, and I leave it to you to show that the latter is impossible: it would unbalance the flow at $p^-$. (The argument here uses the hypothesis that the minimal cut was constructed using the F-F algorithm.) It follows that $p\not\preceq q$, and by symmetry (or a similar argument) $q\not\preceq p$. Thus, $A$ is an antichain.
The last step is to show that the edges contributing capacity towards $c(S,T)$ are preciesly the edges $\langle s,p^-\rangle$ and $\langle p^+,t\rangle$ such that $p\notin A$; this also uses the hypothesis that the minimal cut was constructed using the F-F algorithm, to show that there are no edges of the form $\langle p^-,q^+\rangle$ with $p^-\in S$ and $q^+\in T$.
Once you’ve worked out these details, you’ll have shown that $c(S,T)=|P|-|A|$ and hence that $|A|=|P|-c(S,T)=|P|-|f|=|C|$, as desired.
Best Answer
Whenever you solve the max-flow problem using a residual graph (e.g. using the Ford-Fulkerson algorithm, or using push-relabel), the vertices you can reach from the source in the residual graph are the vertices on one side of the cut.
We can use this idea even if another method is used: starting from the source vertex, explore the graph by traveling forward along edges not at full capacity, and backward along edges with positive flow. The vertices you can reach in this way are exactly the vertices on one side of the cut.