Let $S$ be a set with $n$ elements.
Let $P(S)=\{X\mid X\subseteq S\}$
Let $H\subseteq\mathcal{P}(S)$ (hypergraph with edge set $S$).
Let $H_{|U}=\{U\cap A\mid A\in H\}$
Let $\dim_{\text{VC}}(H)=1+\max\{|U|\mid U\subseteq S\text{ and }H_{|U}=\mathcal{P}(U)\}$
I have to show that $\dim_{\text{VC}}(H)=1\Rightarrow|H|\leq 1$
However, that would mean (unless it's a mean exercise) that there exists $H$ so that $\dim_{\text{VC}}(H)=1$ and $|H|=1$ (otherwise, why '$\leq$' ?)
From the definition of $\dim_{\text{VC}}(H)$ I concluded that $\dim_{\text{VC}}(H)=1$ means that there exists no subset $U$ of $S$ verifying $H_{|U}=\mathcal{P}(U)$.
Let $H=\{\{a_1,a_2,...,a_m\}\},(a_1,...,a_m)\in S^m$; we have $|H|=1$. Now let $U=\{a_1\}$, we have $\mathcal{P}(U)=\{\{a_1\}\}$. Well, $H_{|U}=\{\{a_1\}\}=\mathcal{P}(U)$.
Therefore we have shown that $|H|= 1\Rightarrow\dim_{\text{VC}}(H)>1$. Or have I done an error somewhere ? It seems strange to me.
Consider $S=\{\emptyset \}$. Therefore $P(S)=\{\{\emptyset \}\}$. We choose $H=P(S)$, so obviously $H\subseteq P(S)$ upholds. Furthermore, $|H|=1$, and every $U\subseteq S$ we choose is the empty set itself, therefore $\dim_{\text{VC}}(H)=1$, and you have a case where $|H|=\dim_{\text{VC}}(H)=1$.
Using this a bit more carefully than I worded it and afterwards stating that we assume $S$ to be different than the empty-set, followed by a bit more careful proof such as you gave above should suffice to prove what is needed.