[proofplan]
The argument is a direct $\varepsilon$-$\delta$ verification. Given $\varepsilon > 0$, we set $\delta = \varepsilon / L$ (handling $L = 0$ separately) and show that $d_X(x, y) < \delta$ forces $d_Y(f(x), f(y)) < \varepsilon$ for every $f \in \mathcal{F}$ simultaneously, which is exactly the definition of uniform equicontinuity.
[/proofplan]
[step:Handle the degenerate case $L = 0$ separately]
When $L = 0$, the Lipschitz condition gives $d_Y(f(x), f(y)) \le 0$ for all $x, y \in X$ and all $f \in \mathcal{F}$. Since $d_Y$ is non-negative, this forces $d_Y(f(x), f(y)) = 0$ for every pair, so each $f \in \mathcal{F}$ is constant. In particular, for every $\varepsilon > 0$ and every $\delta > 0$, the implication $d_X(x,y) < \delta \implies d_Y(f(x), f(y)) = 0 < \varepsilon$ holds for all $f \in \mathcal{F}$. The family is uniformly equicontinuous (with any modulus).
[/step]
[step:Choose $\delta = \varepsilon / L$ and verify the uniform equicontinuity condition]
Assume $L > 0$. Fix $\varepsilon > 0$ and define
\begin{align*}
\delta := \frac{\varepsilon}{L}.
\end{align*}
This $\delta$ depends only on $\varepsilon$ and the global Lipschitz constant $L$, not on the choice of $f \in \mathcal{F}$ or the basepoint $x \in X$. Take any $f \in \mathcal{F}$ and any $x, y \in X$ with $d_X(x, y) < \delta$. The uniform Lipschitz hypothesis gives
\begin{align*}
d_Y(f(x), f(y)) \le L \cdot d_X(x, y) < L \cdot \delta = L \cdot \frac{\varepsilon}{L} = \varepsilon.
\end{align*}
Since this bound holds for every $f \in \mathcal{F}$ and every pair $x, y \in X$ with $d_X(x, y) < \delta$, and $\delta$ is independent of both $f$ and $x$, the family $\mathcal{F}$ is uniformly equicontinuous with modulus $\delta = \varepsilon / L$.
[guided]
We need to verify the definition of uniform equicontinuity: for every $\varepsilon > 0$, there exists $\delta > 0$ such that for **all** $f \in \mathcal{F}$ and **all** $x, y \in X$,
\begin{align*}
d_X(x, y) < \delta \implies d_Y(f(x), f(y)) < \varepsilon.
\end{align*}
The key point is that $\delta$ must be independent of both $f$ and $x$. Ordinary continuity of a single Lipschitz function would also yield $\delta = \varepsilon / L$, but there $\delta$ is allowed to depend on $f$. Here, because every $f \in \mathcal{F}$ shares the same Lipschitz constant $L$, the same $\delta$ works across the entire family simultaneously.
Fix $\varepsilon > 0$. Set $\delta := \varepsilon / L > 0$. Take any $f \in \mathcal{F}$ and any $x, y \in X$ with $d_X(x, y) < \delta$. By the hypothesis that $f$ is $L$-Lipschitz, we estimate
\begin{align*}
d_Y(f(x), f(y)) \le L \cdot d_X(x, y) < L \cdot \frac{\varepsilon}{L} = \varepsilon.
\end{align*}
No property of $f$ beyond the uniform bound $L$ was used, so the same $\delta$ serves every member of $\mathcal{F}$. This establishes uniform equicontinuity.
Why is the uniformity in $f$ the essential content here? Without a common Lipschitz constant, individual functions $f_k$ might be $L_k$-Lipschitz with $L_k \to \infty$. In that case, each $f_k$ is individually uniformly continuous, but the moduli $\delta_k = \varepsilon / L_k \to 0$, so no single $\delta$ works for all $k$ and the family fails to be equicontinuous. The shared bound $L$ prevents this degeneration.
[/guided]
[/step]
