Traveling Salesman's Tour Approx Algorithm: is this really a Hamiltonian Path?
You're wrong and the question is also unclearly worded. First, TSP is usually defined in terms of Hamiltonian cycles, not paths. That is, the salesman is in each city exactly once, except for the start point, which he is in twice. Second, as @avakar explains in his first comment, the algorithm given doesn't repeat vertices.

Also, the question seems to be making an unstated assumption. To flesh out @avakar's second comment, suppose you are given cities $1, \dots, n$ where:


$d(i,n)=k$ for $i&lt; n-2$
$d(n-2,n)=d(n-1,n)=2$
$d(n-2,i)=d(n-1,i)=0$ for $i&lt;n$
$d(i,j)=1$ for all other pairs $i$,$j$.


The optimal tours have cost&nbsp;$n$: e.g., $1, 2, \dots, n-2, n, n-1, 1$. However, the given algorithm will produce a tour like $1, n-1, n-2, 2, 3, \dots, n-3, n, 1$ when started at any vertex except&nbsp;$n$.  This has cost $n-5+2k$, which is more than twice the optimum whenever $k&gt;\tfrac12(n+5)$.