I would like to show that the following space is not a surface:
$X$ is made as an identification space of the unit square $Q=\{(x,y)\mid 0\leq x,y\leq1\}$ with the identifications:
$(0,y)\sim(1,y)$ for all $0\leq y \leq1$
$(x,0)\sim(x+\frac{1}{2},0)$ for all $0\leq x \leq \frac{1}{2}$
$(x,1)\sim(x+\frac{1}{2},1)$ for all $0\leq x \leq \frac{1}{2}$
In particular, I would like to find a point on the space which has a neighbourhood that is non-locally Euclidean.
I have so far managed to convince myself that the space cannot be a surface. This is because if we first consider the unit square with the first identification, we get a cylinder, and the second identification then implies that the top-left point of the unit square, as well as the top-centre point both get identified to the top-right point. Likewise, every point between the top-left point and the top-centre point gets identified to a point between the top-centre point and the top-right point. This produces some "overlap". Of course the same applies for the points at the bottom of the square by symmetry.
My difficulty lies in taking, say, the top-centre point, which gets identified to both the top-left and top-right points, and finding a neighborhood about it that is non-locally Euclidean.
I am afraid that my abilities are not quite adequate for me to be able to provide diagrams, but I hope I have made myself understandable verbally.
Note: By surface I mean a compact, connected subspace of $\mathbb{R}^3$ that is locally homeomorphic to open subsets of $\mathbb{R}^2$.
Any help or input would, as always, be highly appreciated.