The most common proof for Floyd-Warshall's algorithm is an induction proof on the outer-most loop, which says
$\delta^k(i,j)=\begin{cases} \min\{\delta^{k-1}(i,j),\delta^{k-1}(i,k)+\delta^{k-1}(k,j)\} & \text{for}\,k\geq 1\\ w(i,j) & \text{for}\,k=0\,\text{(i.e., base case)}, \end{cases}$
where $d^k(i,j)$ is the shortest path from $i$ to $j$ using $k$. Here I present another proof using induction on the length of any shortest path. I have a feeling that the main idea of my proof is the same as above, but I'm not sure if it is correct (missing something important). Here is the proof below:
Proof. (induction on $l$, the lengths of all shortest paths between all pairs of vertices $x_i$ and $x_j$ in the graph)
Base case $l=1$: the shortest path between some pair of vertices $x_i$ and $x_j$ is exactly the edge $(x_i x_j)$. In this case, $d_{ik}+d_{kj}\geq d_{ij}=k((x_i x_j))$ for all $k$ by the fact that $(x_i x_j)$ is the shortest path. The algorithm will not change $d_{ij}$ in this case, so it is correct.
Now, suppose that the algorithm will correctly find the length of all shortest paths of length up to some $c$, and that the shortest path between some pair of vertices $x_i$ and $x_j$ is $(x_i x_{p_1} x_{p_2}\cdots x_{p_c} x_j)$, which has length $c+1$. Now, let $x_{p_m}$ be the vertex in the path that has the biggest index in the distance matrix, and consider the path divided into $(x_i\cdots x_{p_m}\cdots x_j)$; let $P_1=(x_i\cdots x_{p_m})$ and $P_2=(x_{p_m}\cdots x_j)$. For $P_1$, the algorithm will correctly find its length by the inductive hypothesis, since it is of length at most $c-1$ and it is the shortest path between $x_i$ and $x_{p_m}$; similarly the algorithm will correctly find the length of $P_2$, since it is of length at most $c-1$ and it is the shortest path between $x_{p_m}$ and $x_j$. The algorithm will correctly do the comparison $d_{i p_m}+d_{p_m j}<d_{ij}$ when $d_{i p_m}$ and $d_{p_m j}$ are the length of $P_1$ and $P_2$, respectively, and update $d_{ij}$ correctly. This is because $p_m$ has the biggest index in the matrix, meaning that the algorithm would have already correctly found the length of $P_1$ and $P_2$ and stored them in $d_{i p_m}$ and $d_{p_m j}$, since they share intermediate vertices of smaller index and therefore would have been found in previous iterations of the outer-most loop.
P.S. To clarify, this is for a homework problem (as these problems typically are), and I am asking this after the due date. Most large universities in the U.S. nowadays employs graders (I am one myself), which means that my alternate proof may or may not receive a careful look. My point is that I want to know whether I missed something in the proof.
