One-dimensional closed subgroups of $SO(3)$ I want to prove the following statement

Let $H$ be a 1-dimensional closed subgroup of $SO(3)$. Then $H$ has
  exactly one connected component or two connected components.

I have note that:
(1) The lie algebra $\mathfrak h$ is isomorphic to $\mathbb R$.
(2) $SO(3)$ is compact and so the exponential map $\exp:\mathfrak{so}(3)\to SO(3)$ is surjective. 
But I have no idea how to prove the statement. Any idea?

Since $H$ is a subgroup of $\mathrm{SO}(3)$, it acts on the sphere $S^2\subseteq\mathbb{R}^3$ in a natural way. Moreover, $H$ is $1$-dimensional so this action generates a vector field $X$ on $S^2$. By the Hairy Ball Theorem, $X$ vanishes at some $p\in S^2$. 
Lemma. $H$ preserves the set $\{\pm p\}$.
Proof. Since $X_p=0$, the subgroup $H_p=\{h\in H:h\cdot p=p\}$ is $1$-dimensional. But $\mathrm{SO}(3)_p$ is a $1$-dimensional connected group (isomorphic to circle $S^1$) and $H_p$ is a $1$-dimensional closed subgroup of $\mathrm{SO}(3)_p$, so $H_p=\mathrm{SO}(3)_p$. Now, for all $h\in H$ we have $H_{h\cdot p}=hH_ph^{-1}=h\mathrm{SO}(3)_ph^{-1}=\mathrm{SO}(3)_{h\cdot p}$. Thus, $\mathrm{SO}(3)_{h\cdot p}\subseteq H$ for all $h\in H$. Since $\mathrm{SO}(3)_{x}$ and $\mathrm{SO}(3)_{y}$ generate $\mathrm{SO}(3)$ whenever $x,y\in S^2$ are not collinear, we conclude that $h\cdot p=\pm p$ for all $h\in H$. $\square$
The subgroup of $\mathrm{SO}(3)$ which preserves $\{\pm p\}$ has exactly two connected components. It consists of rotations around the line spanned by $p$ together with flipping the line (or in other words, it is the symmetry group of a unit circle in $\mathbb{R}^3$). Explicitly, when $p=(0,0,1)$, this is the group of matrices of the form
$$
\begin{pmatrix}
\cos\theta & \mp\sin\theta & 0 \\ 
\sin\theta & \pm\cos\theta & 0 \\
0 & 0 & \pm 1\end{pmatrix}.$$
Since $H$ is a one-dimensional closed subgroup of that group, it has at most two connected components. It is either $\mathrm{SO}(3)_{\pm p}$ itself, or the $S^1$-subgroup which preserves $p$ (i.e. with $\pm1=1$ above).
