Floyd Warshall's All pair shortest path problem does not evaluate all possible paths
Yes, the path $(4,2,1,2,5)$ is not evaluated.
However this is not a problem.

Since we assume that the graph doesn't contain any negative cycles (this is a requirement for FW), there exists an optimal path for each pair of nodes that doesn't contain any cycles.
Therefore the algorithm only needs to guarantee, that cycle-less path gets evaluated.

By induction you can show that this is always the case.
Let $k$ be the variable in the most outer loop.
Induction hypothesis: After the $k$th iteration the table contains all the shortest paths between each pair that only uses nodes $\le k$ on the path (without start and end node).

At the beginning this is trivially true.
And also the induction step, going from $k$ to $k+1$, is pretty easy to show:
A minimal path using only nodes $\le k + 1$ from $u$ to $v$ either contains only nodes $\le k$, or it contains a minimal path from $u$ to $k+1$ combined with the minimal path from $k+1$ to $v$. (Here we use the fact that the paths contain no cycles. If a path contains cycles, this statement ca be wrong.)
And exactly this case is handled in the algorithms.

Therefore at the end the table contains all minimal paths.