= Solution
For the unit-modulus <prime> values in this question, the <Granville-Soundararajan distance> is
$$
\boxed{D(f,g;X)^2=\sum_{p\leq X}\frac{1-\operatorname{Re}(f(p)\overline{g(p)})}{p}
=\frac12\sum_{p\leq X}\frac{|f(p)-g(p)|^2}{p}.}
$$
The second identity uses $|f(p)|=|g(p)|=1$. Thus it is the <Euclidean distance> between the finite weighted vectors $(f(p)/\sqrt{2p})_{p\leq X}$ and $(g(p)/\sqrt{2p})_{p\leq X}$. Nonnegativity, symmetry and the <triangle inequality> follow from that <norm> representation, and it vanishes exactly when the two prime-value vectors agree. This proves the <prime-restriction metric for pretentious distance>.
\b[On whole <arithmetic functions> at fixed $X$ it is a <pseudometric>.] For example $f=\mathbf1$ and $g=\mu^2$ have identical values at every <prime> and zero distance for every $X$, but different values at four. Thus the literal identity-of-indiscernibles assertion needs the prime-restriction quotient.
Since $\mu(p)=-1$,
$$
D(1,\mu;X)^2=2\sum_{p\leq X}\frac1p.
$$
For completeness, <partial summation> and the prime-counting form of the <Prime number theorem> give
$$
\sum_{p\leq X}\frac1p=\frac{\pi(X)}X+\int_2^X\frac{\pi(u)}{u^2}\,du
=\log\log X+B_1+o(1).
$$
To verify the constant and the error, replace $\pi$ by $\operatorname{Li}+E$ in the first expression. Differentiating $\operatorname{Li}(u)/u$ shows its two contributions combine to $\log\log X$ plus a constant. The error integral converges absolutely because $|E(u)|/u^2\ll e^{-c\sqrt{\log u}}/u$, and $E(X)/X\to0$. This proves the <Mertens second theorem> in the form needed here. In particular the <pretentious distance between one and the Möbius function> satisfies
$$
\boxed{D(1,\mu;X)^2=2\log\log X+2B_1+o(1),\qquad D(1,\mu;X)\sim\sqrt{2\log\log X}.}
$$
For the last request, retain the earlier <prime> normalization $|f(p)|=1$; the proof also works with the weaker assumption $|f(p)|\leq1$. Without a bound on <prime> values, the original definition and the <absolute convergence> of the displayed measure need not apply to an unrestricted <multiplicative arithmetic function>. For example $f(n)=\mu(n)^2n$ is multiplicative and squarefree-supported, but for $X>e$ the displayed measure has infinite positive mass: its <prime> terms alone are $\sum_p p^{-1/\log X}$, which diverges by the <Prime number theorem>.
Put $L=\log X$, $\sigma=1+1/L$, and use the <Fourier transform> convention $\widehat\nu(t)=\int e^{-itu}\,d\nu(u)$. Since $f$ is supported on <squarefree integers>, the defining property of a <multiplicative arithmetic function> gives the absolutely convergent <Euler product>
$$
\widehat\nu(t)=\sum_{n\geq1}\frac{f(n)}{n^{\sigma+it}}
=\prod_p\left(1+\frac{f(p)p^{-it}}{p^\sigma}\right).
$$
The bound on <prime> values gives $|f(n)|\leq1$ for squarefree $n$, so the <total variation norm of a measure> is bounded by $\zeta(\sigma)$; thus $\nu$ is a <finite measure>. Expanding the logarithm of the modulus, with an absolute $O(1)$ remainder since $\sum_pp^{-2\sigma}\ll1$, gives
$$
\log|\widehat\nu(t)|=\sum_p\frac{\operatorname{Re}(f(p)p^{-it})}{p^\sigma}+O(1).
$$
All constants here are uniform in $t$. Moreover
$$
\sum_pp^{-\sigma}=\log\zeta(\sigma)+O(1)=\log L+O(1),
$$
using the pole of the <Riemann zeta function>. The nonnegative deficits $1-\operatorname{Re}(f(p)p^{-it})$ give
$$
\sum_p\frac{1-\operatorname{Re}(f(p)p^{-it})}{p^\sigma}
\geq D(f,n^{it};X)^2-O(1).
$$
Indeed, the replacement of $p^{-\sigma}$ by $1/p$ for $p\leq X$ costs at most
$$
2\sum_{p\leq X}\frac{1-p^{-1/L}}p
\leq\frac2L\sum_{p\leq X}\frac{\log p}p=O(1),
$$
by the <Mertens first theorem>, while discarding the $p>X$ terms costs nothing in this lower bound. Thus the <large Fourier coefficient criterion for pretentiousness> is
$$
|\widehat\nu(t)|\leq C\log X\,e^{-D(f,n^{it};X)^2}.
$$
In particular,
$$
|\widehat\nu(t)|\geq\delta\log X\ \Longrightarrow\
\boxed{D(f,n^{it};X)^2\leq\log(C/\delta)=O_\delta(1).}
$$
The comparison function is the <Archimedean character> $n^{it}$ with this sign because our <Fourier transform> uses $e^{-itu}$.
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