Here's a proof by induction. However, the case $n=1$ is a special case. Say node $s$ is also node $t$. Otherwise, if this is not allowed, then it cannot be proven.
Now the real base case is $n=2$. This is trivial. The graph that satisfies this case consists of two nodes $s$ and $t$, connected together. Note that the edge from $s$ to $t$ is visited last.
For the inductive case, consider that we have the graph of $n$ nodes, and the edge from $s$ to $t$ is visited last. Then we add another node, connect it to $s$ with a cost less than the cost from $s$ to $t$. Thus it will be visited before the edge between $s$ and $t$, so we have preserved that the edge from $s$ to $t$ will be visited last. Further, we have now shown an example for $n+1$, therefore this proves the inductive case. QED