The closed intervals here mean the arcs including endpoints on the circle. I tried to do it by taking the inverse image of those two closed intervals from $S^1$ to its covering space $\mathbb{R}$ and then constructed a homeomorphism of $\mathbb{R}$ that takes these closed intervals to each other. But the issue is that while projecting it back to $S^1$, I am unable to ensure that this will result in a homeomorphism.
Any other approach is also welcome.
Let $e : \mathbb{R} \to S^1, p(t) = e^{it}$, be the standard covering. An interval in $S^1$ is the image $I = e([a,b])$ of an interval $[a,b] \subset \mathbb{R}$ such that $0 < b - a < 2\pi$. The restriction $e_I : [a,b] \to I$ of $e$ is a homeomorphism. Moreover we have $S^1 = I \cup I'$ with $I' = e([b,a+2\pi])$. Note that $I \cap I' = \{ e^{ia}, e^{ib} \}$.
Consider two intervals $I_k = e([a_k,b_k])$. Define "linear" homeomorphisms $u : [a_1,b_1] \to [a_2,b_2], u(t) = a_2 +\frac{b_2-a_2}{b_1-a_1}(t - a_1)$ and $u' : [b_1,a_1+2\pi] \to [b_2,a_2+2\pi], u'(t) = b_2 +\frac{a_2+2\pi-b_2}{a_1+2\pi-b_1}(t - b_1)$. These induce orientation preserving homeomorphisms $U : I_1 \to I_2$ and $U' : I'_1 \to I'_2$. Now $U$ and $U'$ can be pasted together to an orientation preserving homeomorphisms $h : S^1 = I_1 \cup I'_1 \to I_2 \cup I'_2 = S^1$. By construction $h(I_1) = I_2$.