Such a family is equicontinuous, and hence uniformly equicontinuous, since $K$ is compact.
Fix $x\in K$ arbitrarily, and let $\varepsilon > 0$. Since $f$ is continuous and monotonically increasing (not necessarily strictly), there is a $\delta > 0$ such that for every $y \in [x-\delta,x+\delta] \cap K$ we have $f(x) - \varepsilon/4 \leqslant f(y) \leqslant f(x) + \varepsilon/4$.
Let $a = \min [x-\delta,x]\cap K$ and $b = \max [x,x+\delta]\cap K$. Since $f_n \to f$ pointwise, there is an $n_0$ such that $\lvert f_n(a) - f(a)\rvert \leqslant \varepsilon/4$ and $\lvert f_n(b) - f(b)\rvert \leqslant \varepsilon/4$ for all $n \geqslant n_0$.
For $n \geqslant n_0$, we then have $\lvert f_n(y) - f_n(x)\rvert \leqslant f_n(b) - f_n(a) \leqslant (f(x)+2\varepsilon/4) - (f(x) - 2\varepsilon/4) = \varepsilon$ for all $y \in [x-\delta,x+\delta]\cap K$.
For each of the finitely many $n < n_0$, there is a $\delta_n > 0$ such that $\lvert f_n(y) - f_n(x)\rvert \leqslant \varepsilon$ for $y \in [x-\delta_n,x+\delta_n]\cap K$.
Then we have $\lvert f_n(y) - f_n(x)\rvert \leqslant \varepsilon$ for all $y\in [x-\eta,x+\eta]\cap K$ and all $n$ if we choose
$$\eta = \min \{\delta\}\cup \{\delta_n : n < n_0\}.$$
Thus $\{ f_n\}$ is equicontinuous in $x$. Since $x$ was arbitrary, $\{f_n\}$ is equicontinuous on $K$. Since $K$ is compact, $\{f_n\}$ is uniformly equicontinuous on $K$, and therefore the topologies of pointwise convergence and uniform convergence on $K$ coincide on $\{f_n\}$.
It doesn't converge at the point $0$, but for any other point $x\in[0,1]$ there is an $N\in\mathbb{N}$ such that $x>1/N$, so for all $n\geq N$ we have $x>1/n$ and hence $n\chi_{[0,1/n]}(x)=0$. Of course, if $x\notin [0,1]$ we trivially have $n\chi_{[0,1/n]}(x)=0$ for all $n$.
So the example should have omitted the point $0$.
Best Answer
Define
$$f_n(x) = \begin{cases} n^2 x &0 \leq x \leq \frac 1n \\ \frac 1x &\frac 1n \lt x \leq 1. \end{cases}$$
Then $f_n$ converges pointwise to $\frac 1x$ for $x \in (0, 1]$ and $\forall n~f_n(0)=0.$