Let $(E_i,d_i)$ be a metric space. Remember that $\Gamma\subseteq E_2^{E_1}$ is called equicontinuous at $x_0\in E_1$ if $$\sup_{f\in\Gamma}d_2(f(x),f(x_0))\xrightarrow{x\to x_0}0\tag1.$$
How can we show that $(f_n)_{n\in\mathbb N}\subseteq E_2^{E_1}$ is equicontinuous at $x_0$ if and only if $$\limsup_{n\to\infty}d_2f(x),f(x_0))\xrightarrow{x\to x_0}0\tag2?$$ Most probably, we need to assume $(f_n)_{n\in\mathbb N}\subseteq C(E_1,E_2)$ in at least one of the directions, so feel free to assume that.
Regarding the direction "$\Leftarrow$": Let $\varepsilon>0$. By $(2)$, there is a $\delta>0$ with $$\forall x\in B_\delta(x_0):\ell(x):=\limsup_{n\to\infty}d_2f(x),f(x_0))\in\left[0,\frac\varepsilon2\right)\tag3.$$ Now, by definition of the limit superior, there is a $(n_x)_{x\in B_\delta(x_0)}\subseteq\mathbb N_0$ with $$\forall x\in B_\delta(x_0):\forall n\ge n_x:d_2(f_n(x),f_n(x_0))<\ell(x)+\frac\varepsilon2<\varepsilon\tag4.$$ However, this is not enough to conclude $(1)$ ...
So, how can we show the equivalence?