Need Help : Proving polynomials are continuous, without circular reasoning

The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < \epsilon < 1$, in which case $\epsilon^2 < \epsilon$. Or you could use a convergent sequence $x_n$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = a$, and show that

$$\lim_{n \to \infty}f(x_n)g(x_n) = f\left( \lim_{n \to \infty} x_n \right)g\left(\lim_{n \to \infty} x_n\right) = f(a)g(a)$$

You don't need existence of square root. You just need to have that $\forall \epsilon > 0 \ \exists \ \epsilon^\prime > 0$ s. t. ${\epsilon^\prime}^2 < \epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {\epsilon^\prime}^2 < \epsilon$ which gives you what you need anyway. You don't need the bound $\epsilon$ to actually be reachable.

For $\epsilon < 1$, $\epsilon$ itself can serve as $\epsilon^\prime$.