Proving that shortest path distance of adjacent nodes can't differ by more than one Could someone explain this proof to the following question?

Lemma 22.1 from intro to algorithms
Let $G=(V,E)$ be a directed or undirected graph, and let $s\in V$ be any vertex. Then, for any edge $(u,v)\in E$,
$$\mathrm{dist}(s,v)\leq \mathrm{dist}(s,u)+1\,.$$
Proof.
If $u$ is reachable from $s$, then so is $v$. In this case, the shortest path from $s$ to $v$ cannot be longer than the shortest path from $s$ to $u$ followed by the edge $(u,v)$ and thus the inequality holds.

I drew this to visualize the problem but I don't see how what I drew is wrong and how the proof is correct.

If going from $s$ to $u$ has less vertices along that path than going from $s$ to $v$, than the shortest distance of $s$ to $v$ should be greater than going from $s$ to $u$ and $s$ to $u$ to $v$.
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.