A subgroup of SO(4) that is not closed

Subgroup. $\varphi(s+t)=\varphi(s)\varphi(t)$ and $\varphi(-t)=\varphi(t)^{-1}$, so $H$ is a subgroup of the torus $T=\{R_x\oplus R_y:x,y\in\mathbb{R}\}$, which is a closed subgroup of $SO(4)$ (compact: image of $[0,2\pi]^2$ under a continuous map).

Density in $T$. *Step 1: $\{n\sqrt2\bmod1\}$ is dense in $[0,1)$.* Divide $[0,1)$ into $N$ intervals of length $1/N$. Among the $N+1$ numbers $\{k\sqrt2\}$, $k=0,\dots,N$ (fractional parts), two lie in the same interval (pigeonhole): $|\{k\sqrt2\}-\{l\sqrt2\}|<1/N$, so there is $m=k-l\ne0$ with $\|m\sqrt2\|<1/N$ ($\|\cdot\|$: distance to the nearest integer), and $\|m\sqrt2\|\ne0$ since $\sqrt2$ is irrational. The multiples $jm\sqrt2$, $j\in\mathbb{Z}$, move along the circle $\mathbb{R}/\mathbb{Z}$ in steps of size $<1/N$, so they come within $1/N$ of every point.

*Step 2.* Let $(x,y)\in\mathbb{R}^2$ and $\varepsilon>0$. We have $\varphi(t+2\pi n)=R_t\oplus R_{\sqrt2t+2\pi\sqrt2n}$. Take $t=x$. By step 1 there is an integer $n$ with $2\pi\sqrt2n$ within $\varepsilon$ of $y-\sqrt2x$ modulo $2\pi$ (the multiples $n\sqrt2$ are dense modulo $1$, so the multiples $2\pi n\sqrt2$ are dense modulo $2\pi$). Then $\varphi(x+2\pi n)$ is within distance $O(\varepsilon)$ of $R_x\oplus R_y$. So $H$ is dense in $T$.

$H\ne T$. The element $R_0\oplus R_\pi=\mathrm{diag}(1,1,-1,-1)\in T$ is not in $H$: $\varphi(t)=R_0\oplus R_\pi$ requires $t=2\pi m$ and $\sqrt2t=\pi+2\pi l$ for integers $m,l$, i.e. $2\sqrt2m=1+2l$, impossible since the left side is irrational for $m\ne0$ and for $m=0$ it gives $0=1+2l$.

Not closed. If $H$ were closed in $SO(4)$, then, being dense in the closed subset $T$, we would get $T\subset\overline H=H$, contradicting $H\ne T$.

Remark. $H$ is a one-parameter subgroup that "winds" densely on the torus: it is an injective continuous image of $\mathbb{R}$ (the map $\varphi$ is injective) but not a closed embedded curve. This is the standard example showing that Lie subgroups need not be closed.