2-dimensional Lie subgroups of $SO(3)$

It is well know that $SO(3)$ is a 3-dimensional Lie group. My question is $SO(3)$ don't has 2-dimensional Lie subgroups?

No, it does not. If it did, then its Lie algebra would have a $2$-dimensional Lie subalgebra, and this is impossible. Indeed, the Lie algebra of $SO(3)$, is the span (over $\mathbb{R}$) of $x$, $y$ and $z$, which are subject to the following commutation relations:

$[x,y] = z$, and then proceed cyclically in $x$, $y$ and $z$.

If you have $2$ linearly independent elements in this Lie algebra, their commutator will span another element, where they all span the whole Lie algebra of $SO(3)$. This is similar to the cross-product of vectors in $\mathbb{R}^3$.

A $2$-dimensional Lie subgroup would correspond to a 2D Lie subalgebra of $\mathfrak{so}(3)\cong(\mathbb{R}^3,\times)$, but any two linearly independent vectors in $\mathbb{R}^3$ generate all of it using $\times$.