Consider an unoriented unweighted graph $G$. Let $f(a,b)$ denote the maximum flow from $a$ to $b$.
Theorem: In a graph $G$ and its three vertices $a$, $b$, and $c$ we have $f(a,c) \ge \min \big( f(a,b),f(b,c) \big)$.
Proof (wrong): Let $k = \min \big( f(a,b),f(b,c) \big)$. Consider a set of $k$ directed paths $X$ that realize the flow $f(a,b)$ and a set of $k$ paths $Y$ that realize flow $f(b,c)$. Let $Z = X \cup Y$ and observe that this is a set of $k$ paths that realize a $k$ flow from $a$ to $c$, however, it may contain some edges multiple times and hence, is not valid. Take an edge $e \in Z$ that is duplicated. This edge is duplicated because a path $P_A \in A$ and path $P_B \in B$ overlapped on $e$ (and possibly more edges than that). These path $P_A$ go from $a$ to $b$ and path $P_B$ from $b$ to $c$. Consider part of $P_B$ that goes from $e$ to $b$, then part of $P_A$ from $b$ to $e$ – this is a cycle that can be removed from the flow, removing $e$ from the list of edges that are duplicated while not decreasing the flow from $a$ to $c$. Repeat the above cycle removing procedure until there are no edges that are duplicated.
By removing the duplicated edges the above proof changes structure of the graph, hence, after removing the first cycle one may end up with a duplicated edge that does not lead to $b$ at all, and simply continues $c$ – so one cannot use iteration on this argument.
Proof (standard) Let $k = \min \big( f(a,b),f(b,c) \big)$. Consider the maximum flow minimum cut duality so there exists a cut $C_A$ and $C_B$ such that $C_A$ separates $a$ from $b$ and $C_B$ separates $b$ from $c$ and the smaller of these cuts has size at least $k$. Look at a cut $C_C$ from $a$ to $c$, the vertex $b$ is either before or after the cut. Hence $|C_C| \ge |C_A|$ or $|C_C| \ge |C_B|$, therefore, $|C_C| \ge k$.
Proof (interesting): Let $A$ be a set of paths that realize $f(a,b)$ and $B$ be a set of paths that realize $f(b,c)$. Let each $P_A \in A$ be considered a “man”. For $B$ let us flip all the paths in $B$ to create $B’$, let each $P_B \in B’$ be considered a “woman”. We define preferences based on a distance from path’s beginning, i.e., if $P_A$ shares an edge with $P_1$ and $P_2$ of $B$ then $P_A$ prefers $P_1$ iff its (first) shared edge is before the (first) shared edge with $P_2$. Solve the stable marriage problem ↗ for these paths and let a new path be created from concatenating $P_A$ with $P_B$ iff these paths were matched. Notice that no pair of resulting paths can cross as for some $P_A \in A$ and $P_B \in B$ that cross means they cross before their respective overlaps with their matched paths – this is a pair that is not stable, which cannot happen.