Is every connected surface locally symmetric

Question

Is every connected 2 manifold homeomorphic to a space of the form
$$
\Gamma\backslash G/H
$$
where $ G $ is a Lie group, $H $ a subgroup of $ G $, $G/H $ a symmetric space, and $ \Gamma $ a discrete subgroup of $ G $?
I know this is true for the orientable case. In fact the manifold $ G/H $ is holomorphic as is the covering map with fiber $ \Gamma $. One example of a construction like this for a non orientable surface is the projective plane which is a locally symmetric space given as
$$
\mathbb{R}P^2 \cong O(1) \backslash SO(3)/SO(2)
$$
where the $ O(1) $ action on the left is generated by some reflection from $ O(3) $ acting on the symmetric space $ S^2 \cong SO(3)/SO(2) $ (the round sphere).

Answer 1

Credit to Moishe Kohan https://math.stackexchange.com/a/3604136/724711:

The Uniformization Theorem says that if $S$ is a connected Riemannian surface then it is conformally equivalent to a complete Riemannian surface of constant curvature. In particular:

1. Every smooth (topological) connected surface $S$ (oriented or not) admits a complete metric of constant curvature. Equivalently,

2. $S$ is diffeomorphic (homeomorphic) to the quotient $\Gamma\backslash X$, where $X$ is either the unit sphere $ S^2 $ or the Euclidean plane $ E^2 $ or the hyperbolic plane $ H^2 $ and $\Gamma$ is a discrete subgroup of isometries of $X$ acting freely on $X$ (Note that $ S^2,E^2,H^2 $ are exactly the three simply connected symmetric spaces in dimension $ 2 $).

Part 2 does not follow from Part 1, unless you have "complete" in the statement of Part 1.