If I understand correctly, the question is: given that $f_n\to f$ and $f$ is Lipschitz, what additional
assumptions do we need to conclude that $(f_n)$ is a uniformly Lipschitz sequence?
Concerning
it should be a topological property
I remark that Lipschitz-ness is not a topological property, it is a metric property. If we are on a topological space without a metric, the concept of being Lipschitz does not exist.
Here is a metric assumption: the sequence $(f_n)$ converges in the Lipschitz norm
$$
\|f\|_{\rm Lip} = |f(x_0)|+\sup_{x\ne y}\frac{|f(x)-f(y)|}{d(x,y)}
\tag1$$
The definition of norm (1) involves choosing a base point $x_0$ in our metric space; the term $|f(x_0)|$
is necessary so that the constant functions get nonzero norm.
Let's check. Suppose $\|f_n-f\|_{\rm Lip}\to 0$ and $f$ is Lipschitz. Then there is $N$ such that for all $n\ge N$
we have $\|f_n-f\|_{\rm Lip}\le 1$. Hence $\|f_n\|\le 1+\|f\|_{\rm Lip}$, which is a uniform upper bound on the
Lipschitz constants of $f_n$.
Let
$$
h(x)=\left\{\begin{array}{lll}\mathrm{e}^{-1/x^2} & \text{if} & x>0,\\
0 & \text{if} & x\le 0.\end{array}\right.
$$
Then $h\in C^\infty(\mathbb R)$. Then set
$$
j(\boldsymbol{x})=c\,h\big(1-\|\boldsymbol{x}\|^2\big),
$$
where $\boldsymbol{x}\in\mathbb R^n$, and $c>0$, so that $\int_{\mathbb R^n}j(\boldsymbol{x})\,d\boldsymbol{x}=1$. Clearly, $j\ge 0$, $j\in C^\infty(\mathbb R^n)$
and $\,\mathrm{supp}\,j\subset B(0,1)$ - the unit ball.
Next define $j_e(\boldsymbol{x})=\varepsilon^{-n}j(\varepsilon^{-1}\boldsymbol{x})$. Then $\int_{\mathbb R^n}j_\varepsilon(\boldsymbol{x})\,d\boldsymbol{x}=1$ and let the function
$$
f_\varepsilon=f*j_\varepsilon,
$$
i.e.,
$$
f_\varepsilon(\boldsymbol{x})=\int_{\mathbb R^n} f(\boldsymbol{y})\,j_\varepsilon(\boldsymbol{x}-\boldsymbol{y})\,d\boldsymbol{y}=\int_{\mathbb R^n} f(\boldsymbol{x}-\boldsymbol{y})\,j_\varepsilon(\boldsymbol{y})\,d\boldsymbol{y}=
\frac{1}{\varepsilon^n}\int_{B(0,\varepsilon)} f(\boldsymbol{x}-\boldsymbol{y})\,j(\boldsymbol{y}/\varepsilon)\,d\boldsymbol{y}=\frac{1}{\varepsilon^n}\\=\int_{B(0,1)} f(\boldsymbol{x}-\varepsilon\boldsymbol{y})\,j(\boldsymbol{y})\,d\boldsymbol{y}.
$$
Clearly $f_\varepsilon\in C^\infty(\mathbb R^n)$. Next
$$
f_\varepsilon(\boldsymbol{x}_1)-f_\varepsilon(\boldsymbol{x}_2)=
\int_{\mathbb R^n} \big(f(\boldsymbol{x}_1-\boldsymbol{y})-f(\boldsymbol{x}_2-\boldsymbol{y})\big)\,j_\varepsilon(\boldsymbol{y})\,d\boldsymbol{y}
$$
and hence
$$
\lvert\,f_\varepsilon(\boldsymbol{x}_1)-f_\varepsilon(\boldsymbol{x}_2)\rvert\le
\int_{\mathbb R^n} \lvert\, f(\boldsymbol{x}_1-\boldsymbol{y})-f(\boldsymbol{x}_2-\boldsymbol{y})\rvert\,j_\varepsilon(\boldsymbol{y})\,d\boldsymbol{y}\le\|\boldsymbol{x}_1-\boldsymbol{x}_2\|\int_{\mathbb R^n}j_\varepsilon(\boldsymbol{y})\,d\boldsymbol{y}=\|\boldsymbol{x}_1-\boldsymbol{x}_2\|.
$$
Finally
$$
\lvert\,f_\varepsilon(\boldsymbol{x})-f(\boldsymbol{x})\rvert\le
\left|\int_{B(0,1)} \big(f(\boldsymbol{x}-\varepsilon\boldsymbol{y})-f(\boldsymbol{x})\big)\,j(\boldsymbol{y})\,d\boldsymbol{y}.\,\right|\le \cdots\le \varepsilon.
$$
Best Answer
Suppose that $$M = 1 + \sup_{x\in (a,b)}\{|\overline{D}f(x)|,|\underline{D}f(x)|\}$$ and $f$ is continuous.
Fact. Pick $[c,d]\subseteq (a,b)$. Then $$|f(d) - f(c)|\leq M(d-c)$$ Proof. Define $$S = \big\{x\in [c,d]\,\big|\,|f(x) - f(c)|\leq M(x-c)\big\}$$ Clearly $S$ is closed as $f$ is continuous and $c\in S$. Note that for every $x_0\in [c,d]$ there exists $\delta_{x_0}>0$ such that $$\bigg|\frac{f(x)-f(x_0)}{x-x_0}\bigg|< M$$ for all $x\in (x_0-\delta,x_0+\delta)$. We can rewrite it to $$\big|f(x)-f(x_0)\big| \leq M\cdot |x-x_0|$$ Pick $u = \sup S$. If $u < d$, then $$\bigg|f\left(u+\frac{\delta_{u}}{2}\right) - f(c)\bigg| \leq \bigg|f\left(u+\frac{\delta_{u}}{2}\right) - f(u)\bigg| + \bigg|f(u) - f(c)\bigg|\leq M\cdot \frac{\delta_{u}}{2} + M\cdot (u - c) = M\left(u+\frac{\delta_{u}}{2} - c\right)$$ Then $u+\frac{\delta_{u}}{2}\in S$. This is a contradiction with $u = \sup S$. This implies $u = d$ and hence $$|f(d) - f(c)|\leq M\cdot (d-c)$$
Now Fact implies that $f$ is Lipschitz.
Remark.
Pick a sequence $\{x_n\}_{n\in \mathbb{N}}$ of elements of $S$ such that $\lim x_n = x \in [c,d]$. Then $$|f(x_n) - f(c)| \leq M\cdot (x_n-c)$$ By taking limits of both sides of these inequality and by continuity of $f$ we deduce that $$|f(x) - f(c)|\leq M\cdot (x-c)$$
Hence $x\in S$ and this shows that $S$ is closed.