Question
Suppose $\mathbb{E}[|X|^k]$ exists (is finite). Does $\mathbb{E}[|X|^m]$ exist for all $0 \le m \le k$?
Solution
Yes, $\mathbb{E}[|X|^m]$ exists for all $0 \le m \le k$.
Assume $\mathbb{E}[|X|^k] < \infty$. Then $$ \mathbb{E}[|X|^m] = \int_{-\infty}^{\infty} |x|^{m} f_X(x)\,dx = \int_{|x|\le 1} |x|^{m} f_X(x)\,dx + \int_{|x|>1} |x|^{m} f_X(x)\,dx. $$
On the set $|x| \le 1$, we have $|x|^m \le 1$, so $$ \int_{|x|\le 1} |x|^{m} f_X(x)\,dx \le \int_{|x|\le 1} f_X(x)\,dx \le 1. $$
On the set $|x| > 1$ and for $0 \le m \le k$, we have $|x|^m \le |x|^k$, hence $$ \int_{|x|>1} |x|^{m} f_X(x)\,dx \le \int_{|x|>1} |x|^{k} f_X(x)\,dx \le \mathbb{E}[|X|^k] < \infty. $$
Combining these, $$ \mathbb{E}[|X|^m] \le 1 + \mathbb{E}[|X|^k] < \infty. $$
Therefore, $\mathbb{E}[|X|^m]$ exists for all $0 \le m \le k$.