$\require{enclose}\enclose{horizontalstrike}{\rm Greatest}\!$ Least prime factor of $n$ is less than square root of $n$, proof

It is the smallest prime factor that is less than or equal to $\sqrt{n}$, unless $n$ is prime. One proof is as follows: Suppose $n=ab$ and $a$ is the smallest prime factor of $n$, and $n$ is not prime. Since $n$ is not prime, we have $b\ne1$. Since $a$ is the smallest prime factor of $n$, we have $a\le b$. If $a$ were bigger than $\sqrt{n}$, then $b$ would also be bigger than $\sqrt{n}$, so $ab$ would be bigger than $\sqrt{n}\cdot\sqrt{n}$. But $ab=n$.

As stated, what you wrote is false: for example, $5$ is a prime factor of $15$, but the square root of $15$ is less than $4$. Not to mention the fact that if $n$ is prime, then its only prime factor is $n$ itself, certainly larger than $\sqrt{n}$.

What is true is that if $n$ is not prime and not equal to $1$, then it must have a prime factor less than or equal to $\sqrt{n}$.

We can prove this by strong induction: assume the result holds for all $k\lt n$, if $k\gt 1$, then either $k$ is a prime, or it has a prime factor that is no more than $\sqrt{k}$. We wish to prove the same is true for $n$.

If $n$ is prime, we are done. If $n$ is not prime, then there exist $a$ and $b$, such that $1\lt a,b\lt n$ and $n=ab$. We cannot have both $a$ and $b$ greater than $\sqrt{n}$, because then $n = ab \gt \sqrt{n}\sqrt{n} = n$, which is impossible. So either $a\leq\sqrt{n}$, or $b\leq \sqrt{n}$. If $a\leq\sqrt{n}$, then either $a$ is prime, and so $n$ has a prime factor less than or equal to $\sqrt{n}$; or else $a$ has a prime factor $p$ with $p\leq\sqrt{a}$; but a prime factor of $a$ is also a prime factor of $n$, and $a\lt n$ implies $\sqrt{a}\lt\sqrt{n}$, so $p$ is a prime factor of $n$, $p\leq \sqrt{n}$. Either way, $n$ has a prime factor less than or equal to $\sqrt{n}$.

If $b\leq\sqrt{n}$, then repeat the argument with $b$ instead of $a$.