Why is there an unstrict inequality in this proof? (uniform convergence of sequence of functions)

Let $$(X,d_X)$$ and $$(Y,d_Y)$$ be two metric spaces, and let $$\{f_n\}$$ be a sequence of functions $$f_n: X \rightarrow Y$$. For any function $$f : X \rightarrow Y$$ the following are equivalent:

(i) $$\{f_n\}$$ converges uniformly to $$f$$

(ii) $$\sup\{d_Y(f_n(x),f(x)) \ | \ x \in X\} \rightarrow 0$$ as $$n$$ $$\rightarrow \infty$$

Proof: (i) $$\Rightarrow$$ (ii): Assume $$\{f_n\}$$ converges uniformly to $$f$$. $$\forall \epsilon > 0$$ $$\exists N \in \mathbb{N}$$ s.t. $$d_Y(f_n(x),f(x)) < \epsilon$$ $$\forall x \in X$$ and $$\forall n \geq N$$. This means that $$\sup\{d_Y(f_n(x),f(x)) \ | \ x \in X\} \leq \epsilon$$ $$\forall n \geq N$$, and since $$\epsilon$$ is arbitrary, this implies $$\sup\{d_Y(f_n(x),f(x)) | x \in X\} \rightarrow 0$$

I don't understand why there is $$\leq$$ and not $$<$$ in $$\sup\{d_Y(f_n(x),f(x)) \ | \ x \in X\} \leq \epsilon$$. Does it follow from the fact that a supremum is the smallest element greater than or equal to the others?

If $$A\subseteq \mathbb{R}$$ is such that any $$a\in A$$ satisfies $$a then $$\sup A \leq b$$.
The weak inequality is necessary. For instance if $$A=(0,1)$$, then any $$a\in A$$ satisfies $$a<1$$ however $$\sup A = 1$$.
This should explain the weak inequality in your question. However since $$\varepsilon$$ is arbitrary it doesn't really change anything.