Is a prime factor of a number always less than its square root?

No. Consider that the square root of $14$ is about $3.74$ but $14$ has $7$ as a prime factor. Also consider that any prime number such as $2$ is its own (only) prime factor, and any number greater than $1$ is greater than its square root. The theorem you have stated is incorrect: $25$ has no prime factor less than $5$, and $3$ has no prime factor less than $1.732$; however, it is true that every composite number has a prime factor less than or equal to its square root.

You seem to be confused with another statement, which is that the smallest prime factor of a composite number N is less than or equal to $\sqrt N$.

Proof: Suppose $n$ is a positive integer s.t. $n=pq$, where $p$ and $q$ are prime numbers. Assume $p>\sqrt{n}$ and $q>\sqrt{n}$. Multiplying these inequalities we have $p.q>\sqrt{n}.\sqrt{n}$, which implies $pq>n$. This is a contradiction to our hypothesis $n=pq$. Hence we can conclude that either $p\leq \sqrt{n}$ or $q \leq \sqrt{n}$.