Fundamental group of $\{\text{diag}(A,B) \colon A,B \in SO(2)\} /\{I,-I\}$

I could be mistaken here, but it seems that $K\approx SO(2)\times SO(2)\approx S^1\times S^1$ and that the map $K\to K/\{\pm I\}$ gives a fiber sequence

$$\{\pm I\}\hookrightarrow K\to K/\{\pm I\}$$

which we can right more topologically as a fiber sequence

$$S^0\hookrightarrow S^1\times S^1\xrightarrow q K/\{\pm I\}$$

(if you do not know about fibrations, then I am really just saying that $S^1\times S^1$ is a degree $2$ covering space of $K/\{\pm I\}$). Now, we know the univesal cover of $S^1\times S^1$ is $\mathbb R\times\mathbb R\xrightarrow p S^1\times S^1$, so the universal cover of $K/\{\pm I\}$ is $$\mathbb R\times\mathbb R\xrightarrow pS^1\times S^1\xrightarrow qK/\{\pm I\}$$ and we simply need to determine the fiber above a point. It is not hard to see that the fiber above the identity of $K/\{\pm I\}$ is, using more algebraic than topological notation, $$(\mathbb Z\oplus\mathbb Z)\cup(\mathbb Z\oplus\mathbb Z+(1/2,1/2))=\mathbb Z(1/2,1/2)\oplus\mathbb Z(1,0)$$ and so we get that the $\pi_1(K/\{\pm I\})\simeq\mathbb Z\oplus\mathbb Z$ since it is $\pi_0$ of the fiber of $q\circ p$ which is the above discrete group.

It is well-known that:

Any compact, connected abelian Lie group is isomorphic to some $T^k$.

Now because $K$ is compact (why?), connected and abelian (why?), it must be isomorphic to $T^k$ for $k=\dim (K)= 1+1=2$. i.e. $K=\Bbb S^1\times \Bbb S^1$.

Like @JasonDeVito's argument in this post:

$\mathbb{Z}/2\mathbb{Z}\subseteq T^2$ generated by $\langle (\pi, \pi)\rangle$ (note that $-I\sim \tau(x,y):=(x+\pi,y+\pi)$) is normal (since $T^2$ is abelian, so we can form the quotient $K':=T^2/(\mathbb{Z}/2\mathbb{Z})$). Being the continuous homomorphic image of $T^2$, $K'$ must be a compact abelian Lie group, so it must be isomorphic to $T^2$ as a Lie group. In particular, $K'$ is diffeomorphic to $T^2$.

So $\pi_1(K')=\pi_1(T^2)$.