= Solution
Expand the finite <Euler product> into convergent geometric series. Every positive integer $n\le x$ has all its <prime factors> at most $x$, so
$$
P(x)^{-1}=\prod_{p\le x}\sum_{k\ge0}p^{-k}
=\sum_{\substack{n\ge1\\\text{all prime factors of }n\le x}}\frac1n
\ge\sum_{n\le x}\frac1n.
$$
The <harmonic series> diverges, hence \b[$P(x)\to0$]. If the sum of reciprocal primes were finite, then the inequality $-\log(1-t)\le2t$ for $0\le t\le1/2$ would give
$$
-\log P(x)=\sum_{p\le x}-\log(1-1/p)\le2\sum_p\frac1p<\infty,
$$
uniformly in $x$, contradicting $P(x)\to0$. Thus \b[the sum of prime reciprocals diverges]. <Unique factorization> justifies the expansion; no estimate for the distribution of primes is required. This is the <Euler proof that the sum of reciprocals of primes diverges>.
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