Confused with the proof that Edmonds-Karp always monotically increases the shortest-paths
Answer to Question 1:

Because of how we chose , we know that the distance of vertex  from the source  did not decrease, i.e., ′(,)≥(,)
Why the shortest-path from  to  after the augmentation of flow  must be greater than or equal to the distance before the augmentation ? How it is possible to have greater distance that it was ? The augmentation only removes the edge (,). But, the vertex  still in place, isn't it must be strictly equal ?

Because in the Proof, &quot;We will suppose that for some vertex $v \in  V - \left \{  s, t\right \}$, there is a flow augmentation that causes the shortest-path distance from $s$ to $v$ to decrease&quot;. So, if we consider $u$ is the opposite type compared to $v$, that is to say, the shortest-path distance from $s$ to $u$ does not decrease: equation (26.13)  $\delta _{f'}\left ( s,u \right )\geq \delta _{f}\left ( s,u \right )$ holds.
Of course, if you suppose that $u$ is same type as $v$, which has decreasing shortest-path distance from $s$ to $u$, then we can take $u$ to replace the $v$, and find the path, $p:s\rightarrow x \rightarrow u$, where $\delta _{f'}\left ( s,x \right )\geq \delta _{f}\left ( s,x \right )$ holds. In this case $x$ takes the place of $u$ in eq(26.13).
Answer to Question 2:

If we had (,)∈, then we would also have (,)≤(,)+1
This looks like ok for me to have such case. We claim that distance from  to  is less than or equal to the distance from  to  plus 1. But in the graph before the first augmentation this case must be true. It is false for less than case, but true for equal case.

If the shortest path from $s$ to $v$, $p_v: s\rightarrow u \rightarrow v$ overlaps with the shortest path from $s$ to $u$, $p_u: s\rightarrow u$, the equal case happens.
$\delta_f(s,v)=\delta_f(s,u)+1$.
If the two paths $p_v$ and $p_u$ are not overlapped, the shortest path from $s$ to $v$, $p_v: s \rightarrow v$ would not contain the node $u$. Since $p_v$ is the shortest path, obviously the length of $p_v$ should be shorter than the length of $p_u$ + $(u,v)$ (another path from $s$ to $v$). Therefore the less than case happens: $\delta_f(s,v)&lt;\delta_f(s,u)+1$.
Answer to Question 3:

How can we have (,)∉ and (,)∈′ ? The augmentation must have increased the flow from  to . The Edmonds-Karp algorithm always augments flow along shortest paths, and therefore the shortest path from  to  in  has (,) as its last edge.
The edge set $_$ is before the first augmentation and $_{′}$ is after it. The first augmentation must remove the edge (,) and put the reverse edge (,) into the edge set $_{′}$, but not the same edge (,). That is why how we can have (,) in $_{′}$ ?

$(u,v)\notin E_f$ and $(u,v)\in E_{f'}$ mean that the newly augmentation augments flow (a cancellation flow for $(u,v)$) along the edge $(v, u)$. Recall that, augmentation happens on the shortest path, which means that $(v,u)$ is on the shortest path $p$ from $s \leadsto v \rightarrow u \leadsto t$. If $p$ overlaps with the shortest path from $s \leadsto u$, $\delta _{f}\left ( s,v \right ) =  \delta _{f}\left ( s,u \right )-1$ holds. However I am not sure now that the overlapping does happen.