Geodesic is a path

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