You should first make precise what you mean by a (closed/open?) arc and by increasing (decreasing).
Obviously $\operatorname{arc}(x,y)$ is defined for any two points $x = e^{is}, y= e^{it}$ such that $s < t$ and $t -s < 2 \pi$; it is the image of $[s,t] \subset \mathbb{R}$ under the map $e : \mathbb{R} \to S^1, e(x) = e^{ix}$ (if you mean the closed arc). But $e$ is a covering so that $f \circ e : \mathbb{R} \to S^1$ lifts to a map $\tilde{f} : \mathbb{R} \to \mathbb{R}$ such that $e \circ \tilde{f} = f \circ e$. This lift is not unique, but if $\tilde{f_i}$ are any two lifts, we have $\tilde{f_2}(x) = \tilde{f_1}(x) + 2 k \pi$ for some $k \in \mathbb{Z}$.
What you mean by "$f$ is increasing on $\operatorname{arc}(x,y)$" is that $\tilde{f}$ is increasing on $[s,t]$.
With these clarifications your proof is correct, but we can say more. The map $f$ is in fact a covering, see When is a local homeomorphism a covering map?. This implies that $\tilde{f}$ is increasing or decreasing on all of $\mathbb{R}$.
You can see this also directly. W.l.o.g. let $\tilde{f}$ be increasing on some interval $[a_0,b_0]$.
Let $A = \inf \{a \le a_0 \mid \tilde{f} \text{ increasing on } [a,b_0] \}$, $B = \sup \{b \ge b_0 \mid \tilde{f} \text{ increasing on } [a_0,b] \}$. Assume that $B < \infty$. Then $\tilde{f}$ is increasing on $[a_0,B]$. But for some $\epsilon > 0$ the map $\tilde{f}$ must be increasing or decreasing on $[B - \epsilon, B + \epsilon]$. Hence it must be increasing. This contradicts the definition of $B$ and we conclude $B = \infty$. Similarly $A = -\infty$.
