Define $$\omega_f(\delta) =\sup \{|f(x)-f(y)|: (x,y)\in [0,1]^{2}\text{ and } |x-y|\leq \delta\}$$ where $f\in \mathcal{C}([0,1])$ and $\delta\geq 0.$ I have proven that for all $\delta_1,\delta_2\geq 0$ we have that: $$\omega_f(\delta_1)\leq \omega_{f}(\delta_1+\delta_2)\leq \omega_f(\delta_1)+\omega_f(\delta_2).$$ I now want to show that $\omega_f(\delta)$ is continuous for all $\delta\in [0,1].$ I have argued for continuity at $0$ using contradiction and the fact that $\omega_f$ is monotonic. Furthermore, for $r\in (0,1]$ and $h\geq 0$ we have that $$\omega_f(r) \leq \omega_f(r+h)\leq \omega_f(r)+\omega_f (h).$$ If we send $h\to 0$ we get that $\omega_f(r+h)\to \omega_f(r)$ and so $\omega_f$ is right continuous. However, I am not sure how to show $\omega_f$ is left-continuous. I tried the following: $$\omega_f(h)\leq \omega_{f}(h+(r-h))\leq \omega_f(h)+\omega_f(r-h)$$ $$\implies \omega_{f}(r)-\omega_f(h)\leq \omega_f(r-h)$$ and similarily since, $$\omega_f(r-h)\leq \omega_{f}((r-h)+h)\leq \omega_f(r-h)+\omega_f(h)$$ $$\implies \omega_{f}(r-h)\leq \omega_f(r).$$ Thus we have that $$\omega_f(r)-\omega_f(h)\leq \omega_f(r-h)\leq \omega_f(r).$$ Sending $h\to 0$ should give left continuity by squeeze theorem.
Is this reasoning correct?
Monotonicity alone of $\omega_f$ is not enough to yield continuity. You have to use that $f$ is (uniformly) continuous.
A small (maybe pedantic) point of notation: $\omega_f(\delta)$ is a real number, so it cannot be continuous. What you meant was that you wanted to prove that $\omega_f$ was continuous on $[0,1]$, or, if you really want to write $\omega_f(\delta)$, write that $\omega_f(\delta)$ is continuous at every $\delta \in [0,1]$.
Otherwise, it seems correct to me (however, you that $\omega_f(r-h) \leq \omega_f(r)$ is a straight consequence of its monotonicity).