Suppose that for each positive integer $n$, $f_n$ is a continuous function on $\Bbb{R}$ to $\Bbb{R}$. I want to show that if some subset $K$ of $\Bbb{R}$, the sequence $\{f_n\}^{\infty}_{n=1}$ is uniformly convergent on $K$, then $\{f_n\}^{\infty}_{n=1}$ is equicontinuous $K$.
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Let $f_i:K\to \Bbb{R},\;i=1,2,\cdots,\,N$ be continuous. Then, $\{f_i\}^{N}_{i=1}$ is uniformly continuous. Let $\epsilon >0$, then there exists $\delta_i>0$ such that $\forall \,x,y\in K$ with $\left|x-y\right|<\delta_i$, we have
\begin{align}\left|f_i(x)-f_i(y)\right|<\epsilon/3\end{align} Take $\delta=\min\{\delta\}^{N}_{i=1}$. Then, $\forall \,x,y\in K$ with $\left|x-y\right|<\delta$, we have \begin{align}\left|f_i(x)-f_i(y)\right|<\epsilon/3,\forall\,i=1,\cdots,N\end{align} which implies that $\{f_i\}^{N}_{i=1}$ is equicontinuous.
Since $\{f_n\}^{\infty}_{n=1}$ is uniformly convergent on $K$, then $\{f_n\}^{\infty}_{n=1}$ is uniformly Cauchy on $K$. So, there exists $N=N(\epsilon)$ such that $\forall\,m,n\geq N$ \begin{align}\left|f_n(x)-f_m(y)\right|<\epsilon/3\end{align} In particular, for $m=N$, \begin{align}\left|f_n(x)-f_N(y)\right|<\epsilon/3,\;\forall\,n\geq N\end{align} Also, $\{f_i\}^{N}_{i=1}$ is equicontinuous, which implies that $f_N$ is is equicontinuous. So, \begin{align}\left|f_N(x)-f_N(y)\right|<\epsilon/3,\;\text{whenever}\;\left|x-y\right|<\delta\end{align} Hence, for $\left|x-y\right|<\delta$, \begin{align}\left|f_n(x)-f_n(y)\right|&=\left|f_n(x)-f_N(x)+f_N(x)-f_N(y)+f_N(y)-f_n(y)\right|\\&\leq \left|f_n(x)-f_N(x)\right|+\left|f_N(x)-f_N(y)\right|+\left|f_N(y)-f_n(y)\right|\\&< \epsilon/3+\epsilon/3+\epsilon/3=\epsilon\end{align}
Please, I want you to help check if this is correct. Suggestions are highly welcome. If need be, an alternative proof will be highly appeciated if I'm wrong. Thanks!