LetA construction used (repeatedy) in the paper
Edgar, G. A., Centered densities and fractal measures, New York J. Math. 13, 33-87 (2007). ZBL1112.28004.
For more information, see that paper.
We construct a compact metric space $E$ for this purpose.
For $n=1,2,3,\dots$ let $k_n \in \{2,3,4,5,\dots\}$ and $r_n \in (0,1/2)$. More precise properties are to be specified later. Let $T_n$ be a countable dense subset of Hilbert spaceset with $l^2$$k_n$ elements (with the discrete topology). A singleton has Hausdorff dimension Let $$ K_n = k_1 k_2\dots k_n,\qquad R_n = r_1 r_2\dots r_n. $$ so $0$$K_n \nearrow \infty$ and $R_n \searrow 0.$ Also let $R_0 = 1$. Our space is $E = \prod_{n=1}^\infty T_n$ with the product topology. So $E$ is separable, socompact. Define metric $\rho$ on $E$ has Hausdorff dimensionas follows: Let $0$$x = (x_1,x_2,\dots), y=(y_1,y_2,\dots) \in E$. The packing dimension of For $x=y$, define $\rho(x,y) = 0$. For $x \ne y$, let $m \in \mathbb N$ be such that $$ x_j=y_j\quad\text{for }1 \le j \le m,\quad\text{and } x_{m+1} \ne y_{m+1} $$ and define $\rho(x,y) = R_m$.
Then
$$
E \text{ has diameter } R_0 .
$$
$E$ is the same asdisjoint union of $k_1$ closed subsets $E[a_1], a_1 \in T_1$,
$$
E[a_1] = \{(x_1,x_2,\dots) \in E : x_1 = a_1\}
\text{ has diameter } R_1 .
$$
Each set $E[a_1]$ is the packing dimensiondisjoint union of $k_2$ closed subsets
$E[a_1,a_2], a_2 \in T_2$,
$$
E[a_1,a_2] = \{(x_1,x_2,\dots) \in E : x_1 = a_1, x_2=a_2\}
\text{ has diameter } R_2
$$
Continue in this way, so that each $E[a_1,\dots,a_m]$ is the closuredisjoint union
of $\overline{E} = l^2$$k_{m+1}$ closed subsets $E[a_1,\dots,a_m,a_{m+1}], a_{m+1} \in T_{m+1}$, namelyand $\infty$$E[a_1,\dots,a_m]$ has diameter $R_{m}$.
For each $m$, the space $E$ is the disjoint union of $K_m$ closed sets of diameter $R_m$.
HereDefine a "uniform" measure $\mu$ on $T$ so that $$ \mu\big(E[a_1,\dots,a_m]\big) = \frac{1}{K_m} . $$
Now, let $E$$s\in(0,\infty)$. The upper $s$-density of $\mu$ at a point $x \in E$ is separable but not complete $$ \overline{D}_\mu^s(x) = \limsup_{\eta\searrow 0} \frac{\mu(B_\eta(x))}{(\operatorname{diam} B_\eta(x))^s} = \limsup_{n \to \infty} \frac{1/K_n}{R_n^s}= \limsup_{n \to \infty} \frac{1}{K_nR_n^s} . $$ The lower density is $$ \underline{D}_\mu^s(x) = \liminf_{n \to \infty} \frac{1}{K_nR_n^s} . $$ Now what we need to do is select $k_n, r_n$ at the beginning so that, for all $s \in (0,\infty)$ we have $$ \limsup_{n \to \infty} \frac{1}{K_nR_n^s} = +\infty,\qquad \liminf_{n \to \infty} \frac{1}{K_nR_n^s} = 0 $$ Then our metric space $(E,\rho)$ will have Hausdorff dimension $0$ and not compactpacking dimension $\infty$.