= Solution
Let $\phi\in\operatorname{Ham}_c(\mathbb R^{2n})$ be nonidentity and choose $x$ with $\phi(x)\ne x$. Choose disjoint open neighbourhoods $U,V$ of $x,\phi(x)$, respectively. Continuity gives a small Euclidean ball $B=B(x,r_0)$ with $B\subset U$ and $\phi(B)\subset V$. Therefore \b[$\phi(B)\cap B=\varnothing$].
Inside this ball construct an explicit closed <rational Lagrangian submanifold>. For $0<a<r_0/\sqrt n$, take the product <Lagrangian torus>
$$
L=x+\prod_{j=1}^n\{(q_j,p_j):q_j^2+p_j^2=a^2\}.
$$
Its distance from $x$ is $\sqrt n a<r_0$. Each coordinate circle is one-dimensional, so the pullback of $dq_j\wedge dp_j$ to it is zero; summing proves that this $n$-dimensional product is a <Lagrangian submanifold>. It is compact without boundary.
On the $j$th fundamental circle, $q_j=a\cos t$, $p_j=a\sin t$, and the primitive in part (a) restricts to $\tfrac12a^2dt$. Its integral is $\pi a^2$. Translation of the torus adds only an exact constant-coefficient one-form to the primitive, so it does not change the periods. Since $H_1(T^n;\mathbb Z)$ is generated by these circles,
$$
\operatorname{Per}(L)=\pi a^2\mathbb Z,\qquad\gamma(L)=\pi a^2>0.
$$
The given displacement-energy bound applies to this explicitly constructed torus. Because $L\subset B$, the map $\phi$ displaces $L$, and hence
$$
\boxed{\rho_\infty(\mathrm{id},\phi)\geq e(L)\geq\frac{\gamma(L)}2=\frac{\pi a^2}2>0}.
$$
This is <Hofer nondegeneracy from rational Lagrangian displacement>. For two different maps $\phi,\psi$, apply it to $\phi^{-1}\psi\ne\mathrm{id}$ and use the distance definition. Therefore $\rho_\infty(\phi,\psi)>0$. The equivalent oscillation <Hofer metric> is nondegenerate as well.
Only the stated energy inequality was used; its conclusion was not used to presume positive energy of an arbitrary open set. The rational torus and its positive period have been constructed inside the particular displaced ball.
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