Define $$\omega_f(\delta) = \sup \{|f(x)-f(y)|: (x,y)\in [0,1]^{2}\text{ and } |x-y|\leq \delta\}.$$ Let $F(f)(\delta) = \omega_f(\delta)$ for arbitrary $\delta\geq 0$ and $f\in \mathcal{C}([0,1]).$ Then show that $$||F(f)-F(g)||_{\infty}\leq 2||f-g||_{\infty}.$$
Proof Attempt: First we observe that for $\delta\geq 0$ $$\omega_f(\delta) = \max\{|f(x)-f(y)|:(x,y)\in[0,1]^2 \text{ and } |x-y|\leq \delta\}.$$ This is simply because $f$ is continuous function over a compact set and thus there exists $x_1,x_2$ such that $|x_1-x_2|\leq \delta$ and $\omega_f(\delta) = |f(x_1)-f(x_2)|.$ Thus we have the following inequality for $\delta\geq 0,$ $$|F(f)(\delta) - F(g)(\delta)| = | |f(t_1) -f(t_2)| - |g(t_3)-g(t_4)||\leq |f(t_1)-f(t_2) - g(t_3)+g(t_4)|$$ $$\leq |f(t_1)-g(t_3)| +|f(t_2)-g(t_4)|\leq 2||f-g||_{\infty}$$ where $(t_1,t_2), (t_3,t_4)\in [0,1]^2$ with $|t_1-t_2|\leq \delta$ and $|t_3-t_4|\leq \delta$ such that $F(f)(\delta) = \omega_f(\delta) = |f(t_1)-f(t_2)|$ and $F(g)(\delta) = \omega_f(\delta) = |g(t_3)-g(t_4)|.$ Since the above inequality holds for each $\delta\geq 0$ we have that $\newcommand\norm[1]{\left\lVert#1\right\rVert}$ $$\norm{F(f)-F(g)}_{\infty}\leq 2\norm{f-g}_{\infty}.$$
Is this reasoning correct?
Your proof is not correct, because you assume that $$ |f(t_1)-g(t_3)| + |f(t_2)-g(t_4)| \leq 2 \Vert f-g \Vert_{\infty} $$ for arbitrary $t_1, t_2, t_3, t_4$. Choosing $f(x) = g(x) = x$ shows that this can not be true in general.
I would proceed as follows: For $\delta \ge 0$ and all $(x, y) \in [0, 1]^2$ with $|x-y| \le \delta$ $$ |f(x) - f(y)| \le |f(x) - g(x)| + |g(x) - g(y)| + |g(y) - f(y) | \\ \le \Vert f-g \Vert_{\infty} + \omega_g(\delta) + \Vert f-g \Vert_{\infty} $$ which means that the right-hand side is an upper bound for the set $$ \{|f(x)-f(y)|: (x,y)\in [0,1]^{2}\text{ and } |x-y|\leq \delta\} $$ and therefore $$ \omega_f(\delta) \le \omega_g(\delta) + 2 \Vert f-g \Vert_{\infty} \, . $$ The same relationship holds with $f, g$ interchanged, so that $$ |\omega_f(\delta) - \omega_g(\delta)| \le 2 \Vert f-g \Vert_{\infty} \, . $$ This holds for all $\delta \ge 0$, therefore $$ \Vert F(f)-F(g) \Vert_{\infty} \leq 2 \Vert f-g \Vert_{\infty}\, . $$
Remark: We did not use the fact that the functions are continuous, or that the domain is a compact interval. The same conclusion holds for arbitrary bounded functions defined on arbitrary subsets of $\Bbb R$ (or even between metric spaces, if the differences are measured in the corresponding metrics).