The following claim regarding Groups isn't correct, why?

Although $\Bbb Z_3\cong H$ and $\Bbb Z_5\cong K$ for $H,K\le \Bbb Z_{30}$, we have

$$H=\langle 10\rangle\quad\text{and}\quad K=\langle 6\rangle,$$

so that $H\cap K=\{0\}$. This is because $\langle 10\rangle =\{ 0,10,20\}$ and $\langle 6\rangle=\{ 0, 6,12, 18, 24\}$.

For a proof of the statement, see @Riemann'sPointyNose's answer.

The statement is correct. ${H_1\cap H_2}$ is both a subgroup of ${H_1}$ and ${H_2}$, and by Lagrange's theorem ${|H_1\cap H_2|}$ must then have order that divides both ${|H_1|}$ and ${|H_2|}$. Since ${|H_1|,|H_2|}$ are relatively prime, the only way this is possible is if and only if ${|H_1\cap H_2| = 1}$, i.e. ${H_1\cap H_2 = \{0\}}$ is the trivial subgroup.