= Solution
Use the separated-set definition of <topological entropy>. For a <compatible metric> $d$ and $n\geq1$, define the <Bowen metric> and separated-set count by
$$
d_n(x,y)=\max_{0\leq k<n}d(f^kx,f^ky),\qquad
s_n(d,\epsilon)=\max\{|E|:d_n(x,y)\geq\epsilon\text{ for distinct }x,y\in E\}.
$$
For fixed $n$, the <metric> $d_n$ generates the original <topology>: it includes the $k=0$ term and all finitely many <iterations of a map> are <continuous>. Thus its space is compact and $s_n(d,\epsilon)$ is finite. With natural logarithms, set
$$
h_d(f,\epsilon)=\limsup_{n\to\infty}\frac{\log s_n(d,\epsilon)}n,qquad
h_d(f)=\lim_{\epsilon\downarrow0}h_d(f,\epsilon).
$$
The last limit exists, possibly with value infinity, because decreasing $\epsilon$ can only increase separated-set counts.
Let $d'$ be another <compatible metric>. The identity $(X,d)\to(X,d')$ is a <continuous function> on a compact domain, hence <uniformly continuous>. For every $\epsilon>0$, choose $\delta>0$ such that $d(x,y)<\delta$ implies $d'(x,y)<\epsilon$. The same implication at every iterate gives
$$
d_n(x,y)<\delta\ \Longrightarrow\ d'_n(x,y)<\epsilon.
$$
By its contrapositive, every $\epsilon$-<separated set> for $d'_n$ is $\delta$-separated for $d_n$. Consequently $s_n(d',\epsilon)\leq s_n(d,\delta)$ for every $n$, and
$$
h_{d'}(f,\epsilon)\leq h_d(f,\delta)\leq h_d(f).
$$
Taking $\epsilon\downarrow0$ gives $h_{d'}(f)\leq h_d(f)$. Interchanging the two <compatible metrics> gives the reverse inequality. Therefore
$$
\boxed{h_{d'}(f)=h_d(f)=h_{\mathrm{top}}(f).}
$$
<Compactness> is used for <uniform continuity>, not merely for pointwise equivalence of the two <metrics>.
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