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.
I will try to show you why this is intuitive using counter example. Let us call the set reachable by unsaturated paths A, and the other set B. so s is in A, and t is in B. We know that by definition, each edge from A to B is in the minimum cut. Now suppose A were some other set such that it had a node v such that v was not reachable by an unsaturated path. This means that there is an edge such that the path from s to v is saturated, and let its capacity be x and its endpoints be x1 and x2. Also, let the minimum capacity of the edge from v to B be y. If we move x2 to B, and all vertices on x2-v path to B, then we will have a more minimal cut than earlier, since the capacity of our cut is now reduced by y and increased by x (y-x>=0). Therefore it couldnt have been a minimum cut in the first place.
The other case is that some edge from A to B is not saturated. This implies that there is a node in B which is reachable from A by an augmenting path, which again leads to a contradiction.
Best Answer
You can always replace $\infty$ by a very large finite value, so infinite capacity edges cannot cause problems with the max-flow min-cut theorem.
In your particular example, a min-cut is $\{s,a,c,d\}$ on the source side and $\{b,e,t\}$ on the sink side. The cut edges are (s,b), (c,t), (d,t) which have total capacity 3+1+1=5.
Your argument about infinite capacity edges in the cut is not correct because the edges of the flow network (and the cut) are directed. Even though $d$ and $b$ are on opposite sides of the cut, there is no infinite capacity in the cut because $d$ is on the source side and $b$ is on the sink side, whereas in the flow network, the edge is going from $b$ to $d$.