In particular, we want to show that composite numbers all have a prime factor less or equal to $\sqrt n$.
We take a positive composite number n, with a prime factorization $n = p_1 \cdots p_r$.
For the sake of the argument, we assume that $p_1, p_2, \dots, p_r \gt \sqrt n$.
Since $p_i > 1$ we can obtain $p_1 \cdot p_2 \cdot \cdots \cdot p_r \gt (\sqrt n )^r$
If $r = 2k$ then $(\sqrt n)^{2k} = n^k$
If $r = 2k + 1$ then $(\sqrt n)^{2k+1} = (n^k)*(\sqrt n)$
We have $p_1 \cdot p_2 \cdots p_r \gt n^k \gt (n^k)*(\sqrt n)$
This is absurd because $p_1 \cdot p_2 \cdots p_r = n$ and $n \lt n^k$
Therefore, all composite numbers must have at least one prime factor that is less or equal to $\sqrt n$.
More simply, if $n$ is compositie with $r \geq 2$ factors,
$$p_1, p_2, \cdots, p_r > \sqrt n \quad \Longrightarrow \quad n = p_1p_2\cdots p_r > n^{r/2} \geq n$$ i.e., $n > n$, a contradiction. Thus at least one of the prime factors $p_j$ is less than or equal to $\sqrt n$.