You don't need to invoke derivatives, actually. Just write $g_n=f\circ h_n$ where
$$h_n(x) = \max(-n,\min(n,x))$$
Check that $h_n$ is a $1$-Lipschitz function (taking maximum of minimum of Lipschitz functions does not increase the Lipschtiz constant.) The Lipschitz constant is submultiplicative under composition.
Here's a more general statement.
Suppose $K$ is a convex closed subset of $\mathbb R^n$ and $f:K\to \mathbb R$ is an $L$-Lipschitz function. Extend $f$ to $\mathbb R^n$ by $f(x)=f(a)$ where $a$ is the point of $K$ nearest to $x$. Then the extended map is also $L$-Lipschitz.
Proof: let $\pi:\mathbb R^n\to K$ be the nearest point contraction, i.e., $\pi(x)=a$ in the above notation. This map is $1$-Lipschitz (see second half of the answer). Therefore, $f\circ \pi$ is $L$-Lipschitz, and this is our extension. $\quad \Box$.