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I know there are a lot of answers regarding continuity of polynomials. But, this question is different.

We need to have $ lim_xto a x^n = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^th$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_xto a f(x) = L$ and $ lim_xto a g(x) = K$

Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrtepsilon$

And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrtepsilon$

Let $δ=minδ_1,δ_2$

Hence, $0<|x-a|<δ$

⟹$|(f(x)-L)(g(x)-K)-0|<sqrtepsilon sqrtepsilon = ϵ$

Hence, $lim_x to a (f(x)-L)(g(x)-K) = 0$

⟹$lim_x to a (f(x)g(x)-Kf(x)-Lg(x)+KL) = 0$

And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.

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

$$lim_n to inftyf(x_n)g(x_n) = fleft( lim_n to infty x_n right)gleft(lim_n to infty x_nright) = f(a)g(a)$$

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 $epsilon^prime$.