= Solution
Write $Ph=\mathbb E h(X)$ and $P_nh=n^{-1}\sum_{i=1}^nh(X_i)$ for the population and <empirical measure>. The <empirical distribution function> is
$$
\boxed{F_n(t)=\frac1n\sum_{i=1}^n\mathbf1_{\{X_i\le t\}}.}
$$
Fix $\varepsilon>0$ and a finite cover by <function brackets> $[l_r,u_r]$ of $L^1(P)$ width below $\varepsilon$. If $h$ belongs to the $r$th bracket, then
$$
P_nh-Ph\le(P_nu_r-Pu_r)+P(u_r-h)\le|P_nu_r-Pu_r|+\varepsilon,
$$
and, using $l_r$ instead, $Ph-P_nh\le|P_nl_r-Pl_r|+\varepsilon$. Thus
$$
\sup_{h\in\mathcal H}|P_nh-Ph|
\le\varepsilon+\max_r\max\{|P_nl_r-Pl_r|,|P_nu_r-Pu_r|\}.
$$
The <strong law of large numbers> applies to every integrable endpoint because the sample consists of <independent and identically distributed random variables>. For this finite collection the maximum tends to zero on a common probability-one event. Apply the argument for $\varepsilon=1/m$, $m=1,2,\ldots$, and intersect those countably many events. On the resulting event the limsup is at most $1/m$ for every $m$, proving the <uniform strong law from finite L1 bracketing>. This pathwise argument also avoids assuming without justification that an arbitrary uncountable supremum is measurable.
For the <Glivenko-Cantelli theorem>, take $h_t(x)=\mathbf1_{\{x\le t\}}$. Here is a finite bracketing construction which also handles atoms of the distribution. Choose $0<\delta<\varepsilon$ and divide the possible values of $F(t)$ into finitely many bins $[r\delta,(r+1)\delta)$, including the final value one. For each nonempty parameter set $T_r=\{t:r\delta\le F(t)<(r+1)\delta\}$, set $l_r=\inf_{t\in T_r}h_t$ and $u_r=\sup_{t\in T_r}h_t$ pointwise. These functions are indicators of rays, possibly with an open endpoint, so they are measurable. Monotonicity of $h_t$ and monotone limits at the extreme thresholds give $Pl_r=\inf_{t\in T_r}F(t)$ and $Pu_r=\sup_{t\in T_r}F(t)$. Hence $P(u_r-l_r)\le\delta<\varepsilon$. No continuity of $F$ is required. Applying the proved uniform law gives
$$
\boxed{\sup_{t\in\mathbb R}|F_n(t)-F(t)|\longrightarrow0\quad\text{almost surely}.}
$$
For an uncountable class of <continuous functions>, use $\mathcal H=\{h_a(x)=a\sin x:0\le a\le1\}$. The functions are distinct, bounded and integrable for every probability law. Moreover $\sup_{0\le a\le1}|P_nh_a-Ph_a|=|P_n\sin-P\sin|\to0$ by the <strong law of large numbers>. This supplies the requested example without additional compactness machinery.
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