I'm studying the Grassmannian for low dimensions and I saw that $G(2,3)\cong \mathbb{P}_{\mathbb{R}}^2$.
Reference: https://en.wikipedia.org/wiki/Grassmannian
Mostly, I understand the intuitive idea, but I don't know how to formalize that this is true. Could anyone give a rigorous proof of that or recommend me what way should I take?
The general statement is that $G(k,n)$ is homeomorphic to $G(n-k, n)$ for any $0\leq k\leq n$.
One particularly natural homeomorphism $f:G(k,n)\rightarrow G(n-k,n)$ is given as follows. A point $p\in G(k,n)$ is really a $k$-dimensional subspace of $\mathbb{R}^n$. Using the canonical inner product on $\mathbb{R}^n$, there is an orthogonal complement $q$ to $p$. This is an $n-k$ dimensional subspace with $p\oplus q = \mathbb{R}^n$.
Then map $f$ sends $p$ to $q$. The inverse map works the same way.
Applied to your situation, this just says that $G(2,3)\cong G(1,3)$. But, simply note that $G(1,n)$ is the set of $1$-d subspaces in $\mathbb{R}^n$. That is, is the set of all lines though the origin in $\mathbb{R}^n$. So $G(1,n) = \mathbb{P}^{n-1}_\mathbb{R}.$