I'd agree, that's technically a cycle instead of a path. It should be easy to modify the fit the definition exactly though: try the same idea with three nodes, two connecting edges, and a loop.
So what's drawn is an Euler path that is not a Hamiltonian path, but the graph does still have a Hamiltonian path.
Namely, just doing $v_1 \to e_1 \to v_2 \to e_3 \to v_3$. So now you just need to modify it a bit so that visiting every vertex somehow requires you to hit a vertex twice
Hint: think about adding a second "branch" off of $v_1$.
@D.ZackGarza Yes, exactly. Thank you so much @D.ZackGarza! Well, the proof of no existence of HP must include an exhaustive proof, isn't it? Because our graph is finite? Or can we prove using general methods?
@manooooh An exhaustive proof would certainly work here, but it usually suffices to say some words about why one couldn't exist.
And as it turns out, that graph does actually have a Hamiltonian path, so some slight modification is needed.
I think you'd just need to delete $e_4$. Then the proof that it's not Hamiltonian is definitely related to what you said: any path must pass through $v_2$.