How to prove $\sum_p p^{-2} < \frac{1}{2}$?

All primes but 2 are odd numbers so $$\sum_p p^{-2} < 1/4 + \sum_{k=1}^\infty \frac{1}{(2k+1)^2}$$ Using the fact that $1/x^2$ is convex the sum is bounded by $$ \sum_{k=1}^\infty \int_{k-1/2}^{k+1/2}\frac{1}{(2x+1)^2}dx = \int_{1/2}^\infty \frac{1}{(2x+1)^2}dx = 1/4$$

If you know $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ then you could simply say $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} $$ $$\lt \frac{\pi^2}{6} - \frac{1}{1^2}- \frac{1}{4^2}- \frac{1}{6^2}- \frac{1}{8^2}- \frac{1}{9^2}- \frac{1}{10^2}- \frac{1}{12^2}- \frac{1}{14^2}- \frac{1}{15^2}- \frac{1}{16^2} $$ $$ \approx 0.49629 $$ $$ \lt \frac12.$$

Alternatively if you do not know that, instead use $\displaystyle \frac{1}{k^2} \le \int_{x=k-1}^k \frac{1}{x^2}\, dx = \frac{1}{k-1} - \frac{1}{k}$ so $\displaystyle \sum_{n=k}^\infty \frac{1}{n^2} \le \int_{x=k-1}^\infty \frac{1}{x^2}\, dx = \frac{1}{k-1}$ and you can say: $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} \lt \frac{1}{2^2}+ \frac{1}{3^2}+ \frac{1}{5^2}+ \frac{1}{7^2}+ \frac{1}{11^2}+ \frac{1}{13^2}+ \frac{1}{17-1} \approx 0.4982 \lt \frac12.$$

We can deduce this quickly, and without knowing the numerical value of $\pi$, from the fact that $$\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6},$$ for which there are numerous proofs available.

Let $E$ denote the set of even numbers; the sum of the squares of all such numbers is $$\sum_{n \in E} \frac{1}{n^2} = \sum_{k \in \Bbb N} \frac{1}{(2 k)^2} = \frac{1}{4} \sum_{k \in \Bbb N} \frac{1}{k^2} = \frac{1}{4} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{24}.$$ Now, let $X$ denote the union of $\{2\}$ and all positive odd integers $> 1$. In particular, $X$ contains the set $\Bbb P$ of all prime numbers as a subset, and so \begin{align} \sum_{p \in \Bbb P} \frac{1}{p^2} &\leq \sum_{n \in X} \frac{1}{n^2} \\ &= \sum_{n \in \Bbb N} \frac{1}{n^2} - \sum_{n \in E} \frac{1}{n^2} - \frac{1}{1^2} + \frac{1}{2^2} \\ &= \frac{\pi^2}{6} - \frac{\pi^2}{24} - 1 + \frac{1}{4} \\ &= \frac{\pi^2}{8} - \frac{3}{4} . \end{align} So, it suffices to show that $$\frac{\pi^2}{8} - \frac{3}{4} < \frac{1}{2},$$ but rearranging shows that this is equivalent to $\pi^2 < 10$, and $\pi < \frac{22}{7}$ implies $$\pi^2 < \left(\frac{22}{7}\right)^2 = \frac{484}{49} < \frac{490}{49} = 10. $$