Suppose $(X,d)$ complete metric spaces, $D\subseteq X$ dense in X, and $\mathcal{F}=\{f_n:X\to \mathbb{R}: n\in \mathbb{N}\}$ is equicontinuous on $X$. If $\mathcal{F}$ is bounded (above, below) on $D$, then $(f_n)$ converges on $X$.
I wonder that statement above can be proved. My attempt:
Let $\varepsilon>0$. Since $\mathcal{F}$ is equicontinuous on $X$, there exists $\delta>0$ such that for any $y\in X$ with $d(x,y)<\delta$, we have $|f_n(x)-f_n(y)|<\frac{\varepsilon}{4}$ for all $f_n\in \mathcal{F}$.
Then, since $D$ dense in $X$, for $\varepsilon>0$ above and for all $x\in X$, there exists $y_0\in D$ such that $d(x,y_0)<\delta$ and by equicontinuity we have $|f_n(x)-f_n(y_0)|<\frac{\varepsilon}{4}$.
Since $\mathcal{F}$ bounded on $D$, there exists $K>0$ such that $|f_n(x)|\leq K$ for all $x\in D$ and for all $f_n\in\mathcal{F}$. Choose $K=\frac{\varepsilon}{4}$, so we have $|f_n(x)|\leq \frac{\varepsilon}{4}$.
Then, consider for any $m,n\in \mathbb{N}$ with $m,n\geq K$,
\begin{align*}
|f_m(x)-f_n(x)|&\leq |f_m(x)-f_m(y_0)|+|f_m(y_0)-f_n(y_0)|+|f_n(y_0)-f_n(x)|\\
& \leq |f_m(x)-f_m(y_0)|+|f_m(y_0)|+|f_n(y_0)|+|f_n(y_0)-f_n(x)|\\
& < \frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}=\varepsilon.
\end{align*}
so we obtain $(f_n(x))$ is Cauchy sequence on $\mathbb{R}$. Since $\mathbb{R}$ complete, then $f_n(x)$ converges to $f(x)$.
Is there any flaw?
Consider $f_n:[0,1] \to \mathbb{R}$ defined as $f_n(x)=0$ on $[0,1]$ if $n$ is even and $f_n(x)=1$ on $[0,1]$ if $n$ is odd. Then ${\cal F}=\{f_n : n\in \mathbb{N}\}$ is equicontinuous and bounded but $(f_n)$ is not convergent. The flaw lies in "Choose $K= \varepsilon/4,...$". If you have a bound $K$ you can't choose a smaller one, in general.