Proving that union of epsilon nets is an epsilon net

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While reading a paper, I came across these definitions and claims:

Definition: Given $p \in \mathbb R^d$, and $H$, a set of hyperplanes, let $$\text{Violate}_p(H) = \{h \in H: h \text{ is strictly below } p\}$$

Definition: A subset of $R \subseteq H$ is an $\epsilon$-net of $H$, if for every $p \in \mathbb R^d$, $$\text{Violate}_p(R) = \emptyset \implies |\text{Violate}_p(H)| \le \epsilon |H|.$$

Claim: If $R_i$ is an $\epsilon$-net of $H_i$ for each $i$, then $\bigcup_i R_i$ is an $\epsilon$-net of $\bigcup_i H_i$.
That is, $\text{Violate}_p(\bigcup_i R_i) = \emptyset \implies |\text{Violate}_p(\bigcup_i H_i)| \le \epsilon |\bigcup_i H_i|$

My attempt: Fix a point $p \in \mathbb{R^d}$ such that $\text{Violate}_p(\bigcup_i R_i) = \emptyset$.
$\implies \text{Violate}_p(R_i) = \emptyset, \forall i$, since $A \bigcup B = \emptyset \implies A = B = \emptyset$

Now, consider

\begin{align*}\left|\text{Violate}_p\left(\bigcup_i H_i\right)\right| &= |\{h \in \bigcup_i H_i: h \text{ is strictly below } p\}| \\&\le \sum_i |\{h \in H_i: h \text{ is strictly below } p\}| \\&= \sum_i \text{Violate}_p(H_i) \\&\le \sum_i \epsilon |H_i| \end{align*}

Which holds because $\text{Violate}_p(R_i) = \emptyset$ for all $i$.
However, I cannot proceed further becase $|\bigcup_i H_i| \le \sum_i |H_i|$.

How can the claim be proved?