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
$$
\lvert\,f_\varepsilon(\boldsymbol{x})-f(\boldsymbol{x})\rvert\le
\left|\int_{B(0,1)} \big(f(\boldsymbol{x}-\varepsilon\boldsymbol{y})-f(\boldsymbol{x})\big)\,j(\boldsymbol{y})\,d\boldsymbol{y}.\,\right|\le \cdots\le \varepsilon.
$$
Let $\varepsilon>0$ be given, and set $\delta=\min\left[\frac{\varepsilon}{3}, \frac{\varepsilon}{3L}\right]$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \delta) : x \in [a,b] \}$ is a cover for $[a,b]$ we may find a finite subcover, say $\{B(\,x_1, \delta), \, \ldots, \, B(\,x_M, \delta)\}$ (Heine-Borel Theorem). Since $f_n$ converges pointwise on $[a,b]$, for each point $x_j \: \left(\,j=1,\ldots, M \right)$ we may find a positive integer $N_j$ so that
\begin{equation} \left|\, f_n(x_j) -
f_m(x_j) \right| < \frac{\varepsilon}{3} \text{ whenever } n, m \geq N_j \,.
\end{equation}
Setting $N = \max [N_1, \ldots, N_M]$ shows that
\begin{aligned}
\left|\,f_n(x)- f_m(x) \right| & \leq \left| \,f_n (x)- f_n(x_j) \right| + \left|\, f_n (x_j)- f_m(x_j) \right| + \left|\, f_m(x_j)- f_m(x) \right| \\
& < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon
\; \: \text{ whenever } \, n,m \geq N \text{ and } x \in [a,b] .
\end{aligned}
Since $\mathbb{R}$ is complete, it follows that the sequence of functions $\{\,f_n\}_{n=1}^\infty$ converges uniformly on $[a,b]$ (Cauchy Criterion).
I prefer uniform convergence first, ask $\,f$ questions later -_-.
Best Answer
The condition $f \ge 0$ guarantees that $f_n(x)$ is finite for all $x$, and it would suffice to require that $f$ is bounded below.
The example $f(x) = - \Vert x \Vert^2$ shows that without a lower bound, $f_n(x)$ can be identically $-\infty$.
The Lipschitz continuity follows from the triangle inequality: For fixed $x_1, x_2 \in \Bbb R$ and all $y \in \Bbb R$ we have $$ f(y) + nd(x_1, y) \le f(y) + nd(x_2, y) + nd(x_1, x_2) $$ which implies $f_n(x_1) \le f_n(x_2) + nd(x_1, x_2)$. Now exchange $x_1$ and $x_2$ and conclude that $$ |f_n(x_1) - f_n(x_2) | \le n d(x_1, x_2) \, . $$
$f_n(x) \to f(x)$ holds if $f$ is lower semi-continuous at $x$, see for example Lower semicontinuous function as the limit of an increasing sequence of continuous functions.
Without lower semi-continuity it can be wrong. A counter example would the function defined by $f(0) = 1$ and $f(x) = 0$ otherwise, where $0 = f_n(0) \not \to f(0) = 1$.