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 $d$.
While your linear program is a valid formulation of the max flow problem, there is another formulation which makes it easier to identify the dual as the min cut problem.
Let $(G,u,s,t)$ be a network with capacities $u: E(G) \rightarrow \mathbf{R}^+$, source vertex $s$ and sink vertex $t$. Let $P$ be the set of all simple $(s,t)$-paths in $G$. Suppose we have one variable $x_p$ for each possible $p \in P$, which represents how much of the flow is being routed along that path. Then the linear program using these variables becomes
$$ \begin{array}{rcclr}
\text{max} & \sum_{p \in P} x_p & & & \\
\text{subject to} & \sum_{p \ni e} x_p & \leq & u(e) & \forall e \in E \\
& x_p & \ge & 0 & \forall p \in P
\end{array} $$
The first inequalities assure that the capacity of every edge is not violated, and the sum there involves every path containing a certain edge. The second set of inequalities just forces the flow of every path to be non-negative. This formulation has a (possibly) exponential number of variables, but the point here is to reduce the number of constraints, so that the dual becomes easier. Using the flow decomposition you can check that there exists a feasible flow for your LP if and only if there exists a feasible flow with the same cost for my LP, so both formulations are equivalent.
The dual of the new LP has a variable $y_e$ for every edge in $E$, and has the form
$$ \begin{array}{rcclr}
\text{min} & \sum_{e \in E} u(e) y_e & & & \\
\text{subject to} & \sum_{e \in p} y_e & \ge & 1 & \forall p \in P \\
& y_e & \ge & 0 & \forall e \in E
\end{array} $$
The interpretation here is that every edge has a non-negative weight and for every $(s,t)$-path the sum of its edges must be at least $1$. This is a relaxation of the min cut problem. In fact, if you take any $(s,t)$-cut $F \subseteq E$ and consider the characteristic vector $v^F \in \{0,1\}^{E}$ such that $v^F_e = 1$ if and only if $e \in F$; then this vector is feasible for the dual LP with value equal to the value of the cut $F$. So the optimum of the LP is a lower bound for the min cut problem in the network.
Using the duality theorems for linear programming you could prove the max flow min cut theorem if you could prove that the optimum in the dual problem is exactly the min cut for the network, but this needs a little more work. You can check the details in this lecture.
Best Answer
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.