Proving that shortest path distance of adjacent nodes can't differ by more than one
I don't understand your diagram but the point is that, if there is a $x$&ndash;$y$ path of length&nbsp;$k$ then the distance from $x$ to&nbsp;$y$ cannot be bigger than&nbsp;$k$.

So, suppose the shortest path1&nbsp;$P$ from $s$ to&nbsp;$v$ has length&nbsp;$k$. If that path includes&nbsp;$u$, then there is a path of length less than&nbsp;$k$ from $s$ to&nbsp;$u$. Since there is a path of length less than&nbsp;$k$, the distance from $u$ to&nbsp;$v$ is less than&nbsp;$k$, so it is less than or equal to $k+1$, as required.  On the other hand, if $P$&nbsp;does not include&nbsp;$u$, then $P$&nbsp;followed by the edge&nbsp;$uv$ is a path from $s$ to&nbsp;$v$ of length $k+1$. Since there is a path of length $k+1$, the shortest path can't be longer than that, so the distance is at most $k+1$.

1 I should really say "a shortest path", since there can be multiple different paths, all of the same length.