I don't see how the two formulations for uniform convergence are equivalent:
$$ \lim_{n\rightarrow\infty}\sup_{x\in X}\left\{d_Y\left(f\left(x\right){,}\ f_n\left(x\right)\right)\right\}=0 \\ \Leftrightarrow \\ \forall \varepsilon >0:\ \exists N\in \mathbb{N}:\ \forall x\in X:\ n\ge N\ \Longrightarrow \ d_Y\left(f\left(x\right){,}\ f_n\left(x\right)\right)<\varepsilon $$
By the definition of limit, the above is clearly equivalent to
$$ \forall\varepsilon>0:\ \exists N\in\mathbb{N}:\ n\ge N\ \Longrightarrow\ \left|\sup_{x\in X}\left\{d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)\right\}-0\right|<\varepsilon, $$ and since distance is non-negative, $$ \left|\sup_{x\in X}\left\{d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)\right\}-0\right|=\sup_{x\in X}\left\{d_Y\left(f\left(x\right){,}\ f_n\left(x\right)\right)\right\}<\varepsilon.$$
I am not sure however if $$ \sup_{x\in X}\left\{d_Y\left(f_n\left(x\right){,}\ f\left(x\right)\right)\right\}<\varepsilon\ \Leftrightarrow\ d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)<\varepsilon. $$
Could someone clarify?
I'm assuming you meant to write
$$\sup_{x\in X}\left\{d_Y\left(f_n\left(x\right){,}\ f\left(x\right)\right)\right\}<\varepsilon\ \Leftrightarrow\ \forall x \in X, d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)<\varepsilon,$$
since that would facilitate the proof.
The reverse implication does not hold with strict inequality, but this presents no major problem.
It should be if $d_Y(f(x), f_n(x)) < \epsilon$ for all $x \in X$, then $\sup\{d_Y(f(x), f_n(x)): x \in X\} \leqslant \epsilon$.
To prove this, suppose $\sup\{d_Y(f(x), f_n(x)): x \in X\} = \alpha > \epsilon$ and take $\beta$ such that $\epsilon < \beta < \alpha$. There must exist $x' \in X$ such that $d_Y(f(x'),f_n(x')) > \beta> \epsilon$, otherwise $\beta$ is an upper bound less than the least upper bound $\alpha$. However, this contradicts the hypothesis that $d_Y(f(x),f_n(x)) < \epsilon$ for all $x \in X$.
The forward implication that $\sup\{d_Y(f(x), f_n(x)): x \in X\} < \epsilon$ implies $d_Y(f(x), f_n(x)) < \epsilon$ holds as you have written and is straightforward to prove.