The space of oriented 2-dimensional subspaces of $\mathbb{R}^4$ is compact and connected.

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I am trying to

Show that the set of oriented 2-dimensional subspaces of $\mathbb{R}^4$ is 4 dimensional compact and connected homogeneous space.

The given hint is that

(1) $SO(4)$ is smooth transitive Lie group action on the space

and that

(2) each stabilizer of an element in the set is closed subgroup of $SO(4)$.

It is clear that the space is homogeneous due to the hint (1). Now I should verify if the space is compact and connected.

I cannot come up with any idea regarding this problem. I will be very thankful for any help on this problem.

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For compactness: consider the set $$A=\{(u,v):u,v\in\Bbb R^4,\|u\|=\|v\|=1,u\cdot v=0\}.$$ Prove that $A$ is compact. There's a natural continuous map from $A$ onto your space sending $(u,v)$ to the plane they span, with orientation defined by the order of $u$ and $v$. So your space is compact.

Come to think of it, $A$ is connected too. It's a fibre bundle with base $S^3$ and fibres $S^2$.