Let $A$ be a non-empty subset of $\mathbb{R}^n$ and $f : A \to \mathbb{R}$ is be a bounded function. Let $\alpha , \beta \in \mathbb{R}$ be such that $\alpha \leq f( \vec{x}) \leq \beta$ for every $\vec{x} \in A$. Suppose that $u : [ \alpha , \beta ] \to \mathbb{R}$ is a Lipschitz function, and consider the function $g := u \circ f$. I want to prove that for every non-empty set $B \subseteq A$ one has the inequality $\mbox{osc}_B (g) \leq c \cdot \mbox{osc}_B (f)$, where $c$ is a Lipshitz constant for $u$.
My approach:
$$\textbf{(*) }|g(\vec{x})-g(\vec{y})|=|u(f(\vec{x}))-u(f(\vec{y}))|\le c |f(\vec{x})-f(\vec{y})| \text{ (since $u$ is Lipschitz on $B\subset A$)}$$
Hence, $$\textbf{(**) }\mbox{osc}_B (g) = \sup\{|g(\vec{x})-g(\vec{y})|:\vec{x},\vec{y}\in B\}\le c\sup\{ |f(\vec{x})-f(\vec{y})|: \vec{x}, \vec{y} \in B\} = c\cdot\mbox{osc}_B(f).$$
My doubt is whether or not one can claim that $\textbf{(*)}$ implies $\textbf{(**)}$, since the Lipschitz condition assumes that $\textbf{(*)}$ holds for the same $\vec{x},\vec{y}$, whereas in $\textbf{(**)}$ $\vec{x},\vec{y}$ are generally different in the inequality?
Your proof is correct. Indeed, for a given $(\vec{x},\vec{y})\in B^2$, one has
$$\lvert g(\vec{x})-g(\vec{y})\rvert \leq c \lvert f(\vec{x})-f(\vec{y})\rvert \leq c \cdot \operatorname{osc}_B(f), $$ and then taking the sup on all values of $(\vec{x},\vec{y})$ gives
$$\operatorname{osc}_B(g) \leq c \cdot \operatorname{osc}_B(f). $$