Removable singularity of metric?

It's not quite so easy to prove the general case $phi in (0, alpha)$ (the singularity can't be removed in general), but fortunately, for the case $alpha = 2pi$, it is fairly easy.

First let's write our metric in a more appropriate form. This spacetime can be described with two coordinate patches, $(U_1, f_1)$ and $(U_2, f_2)$, with

begin{eqnarray} f(U_1) &=& (0,infty) times (0, 2pi) f(U_2) &=& (0,infty) times (-varepsilon, varepsilon) end{eqnarray}

$U_1$ and $U_2$ overlap in two subsets, both $(0, varepsilon)$ overlap, with the identity as a transition function, and $(2pi - varepsilon, 2pi)$ overlaps with $(-varepsilon, 0)$, with the transition function $phi_2 = 2pi + phi_1$.

On each patch, the metric is indeed just

begin{equation} ds^2 = dr^2 + r^2 dphi^2 end{equation}

That's an entirely fine spacetime , but you'll notice that the radial curve $gamma(l) = (R - l, 0)$ is inextendible beyond $l = R$ and has finite half-length. Therefore we have a singularity at $r = 0$. But fortunately, this spacetime is extendible, which means that there exists a larger spacetime $M'$ such that we have an inclusion map $iota : M to M'$ where $iota$ is an isometry.

We don't need to look too deep for this extended spacetime, picking Cartesian coordinates will do. Then we have the inclusion map (in coordinate form)

begin{eqnarray} iota(r, phi_1) &=& (r cos phi_1, r sin phi_1) iota(r, phi_2) &=& left. begin{cases} (r cos phi_2, r sin phi_2) & text{for } phi_2 in (0, varepsilon) (r cos (2pi + phi_2), r sin (2pi + phi_2)) & text{for } phi_2 in (-varepsilon, 0) end{cases} right. end{eqnarray}

which is also isometric to the Euclidian metric. This will only work for $alpha = 2pi$, as you can notice. You can always map one of the coordinate patch to $mathbb{R}^2$ (it's always locally extendible), but if your total angle isn't $2pi$, the inclusion map will not work as you will not have $cos(phi_1) = cos(alpha + phi_2)$ at the appropriate points.