Let $f$ be the pointwise limit of $f_n$. We want to prove that $\|f-f_n\|_{\infty} \to 0$.
All $f_n$ are $K-$Lipschitz, that means :
$\forall x,y \in [a,b], \forall n, |f_n(x)-f_n(y)|\leqslant K|x-y|$
Take the limit with $n$ : you have that $|f(x) - f(y)| \leqslant K|x-y|$ so that $f$ is $K-$Lipschitz, thus continuous, and for all $n$, $f-f_n$ is continuous.
As $[a,b]$ is a compact, $M_n=||f-f_n||_{\infty} <\infty$. We want to prove that $M_n \to 0$. By continuity, there exists $x_n \in [a,b]$ such that $M_n = |f(x_n)-f_n(x_n)|$.
Moreover, $M_n \leqslant ||f||_{\infty} + ||f_n||_{\infty}$ by triange inequality, and $||f_n||_{\infty} \leqslant |f_n(a)| + K|b-a|$. As $f_n(a)$ is a convergent sequence, $||f_n||_{\infty}$ is bounded, so is $M_n$.
Right now, we know that $M_n$ is a bounded sequence in $\mathbb{R}_+$. To show it converges to $0$, we only have to show that every converging subsequence has limit $0$. Take $M'_n = M_{\varphi(n)}$ a converging subsequence, $f'_n = f_{\varphi(n)}$ and $x'_n = x_{\varphi(n)}$. To show that, use the fact that $[a,b]$ is compact, thus, $x'_n$ has a converging subsequence $x'_{\psi(n)}$ with limit $x_{\infty}$, and use :
\begin{align}
M'_{\psi(n)} &= |f(x'_{\psi(n)}) - f'_{\psi(n)}(x_{\psi(n)})| \\&\leqslant |f(x'_{\psi(n)})-f(x_{\infty})| + |f(x_{\infty}) - f'_{\psi(n)}(x_{\infty})|+|f'_{\psi(n)}(x_{\infty}) - f'_{\psi(n)}(x'_{\psi(n)})| \\
& \leqslant K|x'_{\psi(n)}-x_{\infty}| + |f(x_{\infty}) - f(x'_{\psi(n)})| + K|x'_{\psi(n)} - x_{\infty}|
\end{align}
And all three terms are going to zero because of the hypothesis. As $M'_n$ is convergent, its limit is $0$, and every converging subsequence of $M_n$ has limit $0$. $M_n$ is then going to $0$, which means that $f_n$ converges uniformly.
Sorry, there is many notations, but it's the detailed proof.
