Step 1: Setup. Let $\mathcal{A} = \{n(n+2) : n \leq x\}$, $\mathcal{P} = \{p : p \text{ prime}\}$, $z = x^{1/2}$. An element $n$ survives the sieve (i.e., $(n(n+2), P(z)) = 1$) only if both $n$ and $n + 2$ have no prime factor below $z$, hence both are prime (or 1).

Step 2: Brun's combinatorial sieve. Truncate the Mobius inclusion-exclusion at level $2r$ (even, for an upper bound): $S(\mathcal{A}, z) \leq \sum_{\substack{d | P(z) \\ \Omega(d) \leq 2r}} \mu(d) |\mathcal{A}_d|$ where $\mathcal{A}_d = \{n \leq x : d | n(n+2)\}$.

Step 3: Counting $|\mathcal{A}_d|$. For squarefree $d$ with $d | n(n+2)$: the number of residues $n \bmod d$ with $d | n(n+2)$ is $\omega(d) = \prod_{p|d} \omega(p)$ where $\omega(p) = 2$ for $p > 2$ (since $n \equiv 0$ or $n \equiv -2 \bmod p$) and $\omega(2) = 1$. So $|\mathcal{A}_d| = \frac{\omega(d)}{d}x + r_d$ with $|r_d| \leq \omega(d) \leq 2^{\Omega(d)}$.

Step 4: Main term estimate. The main term gives: $S \leq x \sum_{\substack{d|P(z) \\ \Omega(d) \leq 2r}} \frac{\mu(d)\omega(d)}{d} + \sum_{\substack{d|P(z) \\ \Omega(d) \leq 2r}} 2^{\Omega(d)}$

Choosing $r = c\log\log x$ (Brun's choice), the main term is $\ll x \prod_{2 < p < z}(1 - 2/p) \ll \frac{x}{(\log z)^2} \ll \frac{x}{(\log x)^2}$ times factors of $(\log\log x)^{O(1)}$ from the truncation.

The remainder sum is bounded by $\sum_{d \leq D} 2^{\Omega(d)} \leq D \cdot (\log D)^{O(1)}$ with $D = z^{2r} = x^{c'\log\log x}$, which is $o(x/(\log x)^2)$ for appropriate $c$.

Step 5: Conclusion. $\pi_2(x) \leq S(\mathcal{A}, z) \ll \frac{x(\log\log x)^2}{(\log x)^2}$. The sum $\sum_{p \leq x, p+2 \text{ prime}} 1/p$ is then bounded by partial summation: $\sum 1/p \leq \int_2^x \frac{d\pi_2(t)}{t} \ll \int \frac{(\log\log t)^2}{t(\log t)^2}dt < \infty$. $\square$