Let G a graph (connected) and the walk $T = (u, x_1, \dots , x_k, v)$ a geodesic (walk with the shortest distance) between the vertex $u$ and $v$. Show that: 1. $T$ is a path. 2. If $i,j \in \{1, \dots, k \}$ and $i \leq j$ then the walk $(x_i, T, x_j)$ is a geodesic between $x_i$ and $x_j$.
In (1) I have the idea to led a contradiction: Proof: Suppose that $T$ is not a path, i.e. there's at least one vertex with more than one occurrence. Since $G$ is connected then there's a path between $u$ and $v$. So, let $T'$ a path between $u$ and $v$, since $T$ is a path then it has the shortest distance between these two vertex. So $T'$ is geodesic and $T$ is not. Contradiction.
In (2) I must to assume that the vertex are not adjacent, but since then, what?
For (1), if you remove the first and last occurrence of the repeated vertex in $T$ you obtain a shorter walk Than $T$ which is a contradiction.
For (2), if $(x_i, T, x_j)$ is not geodesic then there is a shorter walk between $x_i$ and $x_j$ which means that $T$ is not geodesic because you can get shorter walk between $u$ and $v$ by replacing $(x_i, T, x_j)$ in $T$ the shorter walk between $x_i$ and $x_j$ which is contradiction