No $n$-stair graph has a Hamiltonian path for any $n \ge 3$. $\textbf{Definition 1:}$ An $n$-stair is the collection of the unit squares bounded by the $x$-axis, $y=x$ and $x=n+1$.
$\textbf{Definition 2:}$ An $n$-stair graph is obtained by replacing each of the unit squares by a vertex where two vertices are adjacent if two unit squares share a common side.
$\textbf{Claim:}$ No $n$-stair graph has a Hamiltonian path for any $n \ge 3$.
I have a kind of visualization of the graph in mind but not getting any clue how to proceed.

Here is a depiction of the $5$-stair graph. Each non-space character is a vertex and there is an edge between and only between adjacent characters:
    e
   xo
  xoo
 xooo
eoooo

Clearly, any Hamiltonian path must start and end at the es, which are only of degree $1$. The path must also pass through the x vertices on the diagonal; since they are degree $2$ their two incident edges must be in the path. When you link up all these "forced" edges, however, you find that a complete path has been formed without any possibility of reaching the other vertices. Thus the $n$-stair graph has no Hamiltonian cycle for $n\ge3$.
