Let $f_n\colon \mathbb{R} \rightarrow \mathbb{R}$ be differentiable for each $n \in \mathbb{N}$ with $\lvert f'_n(x)\rvert \le 1$ for all $n$ and $x$. Let $\lim_{n \rightarrow \infty} f_n(x) = g(x)$ for all $x$. Prove that $g\colon \mathbb{R} \rightarrow \mathbb{R}$ is Lipschitz-continuous.
I know one can prove this using the triangle inequality, e.g. as suggested here. I wanted to check if the following, similar argument is also valid:
Let $x$,$y \in \mathbb{R}$ and $n$ be arbitrary. By the mean value theorem, there exists $c \in (x,y)$ such that $$ f_n(x)-f_n(y)= f_n^{'}(c) \cdot (x-y).$$ Using this, we obtain a bound for the differential quotient of $f_n(x)$: $$\frac{\mid f_n(x)-f_n(y)\mid}{\mid x-y \mid} = \frac{\mid f_n^{'}(c)\mid \cdot\mid x-y \mid}{\mid x-y \mid} \leq 1.$$ Therefore, we obtain the following: $$\frac{\mid g(x)-g(y)\mid}{\mid x-y \mid} = \frac{\mid \lim f_n(x)- \lim f_n(y)\mid}{\mid x-y \mid} =\frac{\mid \lim (f_n(x)-f_n(y))\mid}{\mid x-y \mid} = \lim\frac{\mid f_n(x)-f_n(y)\mid}{\mid x-y \mid} \leq 1 ,$$ where the first equality follows by definition, the second by limit laws for addition, the third as $\mid {}\cdot{}\mid$ is continuous and the inequality as if a convergent sequence is bounded, then the limit is bounded in the same way.