We have the generalization:
Let $(Y,d)$ a metric space and $X$ a non-empsty subset of $Y$ and $\{f_n\}$ a sequence of continuous functions from $X$ to $\mathbb R$. If for all $x_0\in X$, we can find a ball $B:=B(x_0,\varepsilon_{x_0})$ such that the sequence $\{f_n\}$ converges uniformly to $f$ on $B\cap X$, then the family $\{f_n\}$ is equicontinuous.
We fix $x_0\in X$ and $\varepsilon>0$. Let $\delta_1$ such that the sequence $\{f_n\}$ converges uniformly to $f$ on $X\cap B(x_0,\delta_1)$. Let $N$ an integer such that $\sup_{x\in B(x_0,\delta_1)}|f(x)-f_n(x)|\leq\frac{\varepsilon}3$. For each $i\in\{1,\ldots,N\}$ we can find a $\eta_i$ such that if $d(x,x_0)\leq \eta_i$ then $|f_i(x)-f_i(x_0)|\leq \frac{\varepsilon}3$. Put $\delta_2:=\min\{\eta_i,1\leq i\leq N\}$. Finally, since $f$ is continuous at $x_0$ as a uniform limit of such functions, let $\delta_3$ such that if $d(x,x_0)\leq \delta_3$ then $|f(x)-f(x_0)|\leq\frac{\varepsilon}3$. Now put $\delta:=\min(\delta_1,\delta_2,\delta_3)$. Let $x\in B(x_0,\delta)$. Let $n\in\mathbb N$. If $n\leq N$ we are done since $\eta_n\leq\delta$, and if $n>N$ we have \begin{align}|f_n(x)-f_n(x_0)|&\leq |f_n(x)-f(x)|+|f(x)-f(x_0)|+|f(x_0)-f_n(x_0)|\\&\leq 2\sup_{y\in B(x_0,\delta)}|f_n(y)-f(y)|+|f(x)-f(x_0)|\\ &\leq 3\frac{\varepsilon}3, \end{align} hence the family $\{f\}\cup\{f_n,n\geq 1\}$ is equicontinuous at each point of $X$.