= Solution
Expand the finite <Euler product> into <geometric series>. By <unique prime factorization>, it is the sum of $1/n$ over positive <integers> whose <prime factors> do not exceed $x$, and hence contains every term up to $m=\lfloor x\rfloor$. The <harmonic sum> satisfies
$$
\prod_{p\leq x}(1-p^{-1})^{-1}\geq\sum_{n=1}^m\frac1n>\int_1^{m+1}\frac{dt}{t}=\log(m+1)>\log x.
$$
The last inequality uses $m+1>x$, including when $x$ is an <integer>. For each <prime number> $p$,
$$
0<-\log(1-p^{-1})-p^{-1}=\sum_{j=2}^\infty\frac1{jp^j}<\frac1{2p(p-1)}.
$$
The <telescoping series> over all <integers> $n\geq2$ bounds the sum of these remainders by $\frac12\sum_{n\geq2}[n(n-1)]^{-1}=1/2$. Taking <logarithms> of the product bound therefore gives
$$
\boxed{\sum_{p\leq x}\frac1p>\log\log x-\frac12,\qquad c=-\frac12.}
$$
This explicit constant suffices; no asymptotic theorem about the distribution of <prime numbers> is needed.
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