Prove the following claim on Hamilton Path? I am trying to prove the following claim:

Given DAG graph, there is Hamilton path iff the following algorithm
returns true:


Do topologic sorting.
Move on the graph's vertices one by one (from low to high). In case there is no edge connecting 2 vertices with adjacent values from the topologic sorting then return false.
if no false was returned after we check all vertices, return true.


I am stuck of proving one side which is: if there is Hamilton path then the algorithm returns true.
I tried using induction on number of vertices in the graph n:

Base case is simple for n==0.

Assuming claim is correct for n I want to prove for n+1


So I said, let's exclude the last vertex in the given Hamilton path (let's call it a), and assume by contradiction that the algorithm returned false.
This means one of the 2:

Two vertices with adjacent values had no edge connecting them and both aren't a.
this contradictc the assumtion that the claim hold for graph with n vertices.

One of the two vertices is a and the other isn't a.


I am stuck on proving that case (2) will give us a contradiction, How may I continue?
Try to prove instead (without induction) that if the algorithm returned false, there is no hamilton path.
Hint: there must be two parallel nodes. Can we reach one from the other?