$\displaystyle d(f,g)=\sup_{a\leq x \leq b} |f(x)-g(x)|+ \sup_{a\leq x \leq b} |f'(x)-g'(x)|, \quad f,g \in X$.
Approach:
$|f(x)-g(x)| \leq |f(x)-h(x)|+|h(x)-g(x)| \implies$
$ |f(x)-g(x)| \leq \displaystyle\sup_{ x\in [a,b]}\bigg[|f(x)-h(x)|+|g(x)-h(x)| \bigg] \leq \displaystyle\sup_{ x\in [a,b]}|f(x)-h(x)|+ \displaystyle\sup_{ x\in [a,b]}|h(x)-g(x)| \implies$
$\displaystyle\sup_{ x\in [a,b]} |f(x)-g(x)| \leq \displaystyle\sup_{ x\in [a,b]}|f(x)-h(x)|+ \displaystyle\sup_{ x\in [a,b]}|h(x)-g(x)|. \quad \quad $ $(1)$ Similarly, the other inequality $(2)$ is obtained and $(1)+(2)$ gives us the required triangle inequality. [Continuity of the derivative ensures that the function is everywhere defined and bounded]
Is this a valid approach? Please verify.