Intermediate Value theorem, $nth$ root function and continuity 
So this problem is.. ridiculous to be honest. I have no idea where to start or what to do. Any help is appreciated. For the record, I am using the metric spaces definition of continuity.
 A: First we start with showing that $f(x) = x^{n}$ is continuous for all $x$ (actually we need only non-negative values of $x$). OP should be able to prove this by using the fact that $x^{n} = x \cdot x \cdots x\text{ (}n\text{ times})$ where $n$ is fixed positive integer and then using product rule of limits.
Next we need to show that for any $a > 0$ there is a unique $n^{\text{th}}$ root of $a$. What this means is that we have to show the existence and uniqueness of a number $c > 0$ such that $c^{n} = a$. First we show that $c$ (if it exists) is unique. How?? Suppose there are two such values $c, d$ such that $c^{n} = a = d^{n}$ and $c \neq d$. Now use $$0 = c^{n} - d^{n} = (c - d)(c^{n - 1} + c^{n - 2}d + \cdots + cd^{n - 2} + d^{n - 1})$$ and you can see the obvious contradiction here.
Now we need to see the existence of $c$. If $a = 1$ then obvious $c = 1$. If $0 < a < 1$ then we know that $0 = f(0) < a < f(1) = 1$. If $a > 1$ then $f(a) > a$ so that $f(0) < a < f(a)$. Thus in any case we have a relation of the form $f(0) < a < f(b)$ for a suitable value of $b$. Now use IVT for continuous function $f(x)$.
First part of the problem is now established. The second part deals with showing that $f(x) = x^{n}$ is invertible in the domain $[0, \infty)$ and the inverse function, namely the $n^{\text{th}}$ root function is also continuous. For existence of inverse functions what do we need? We need to show that the original function $f(x)$ is a bijection (one-one and onto). From first part of the question itself you should be able to show that $f(x) = x^{n}$ is one-one (this is same logic we used for uniqueness of $c$) and onto (this is same logic we used for existence of $c$ above).
So the inverse function $g(x)$ for $x \geq 0$ exists such that $f(g(x)) = x = g(f(x))$ for all $x \geq 0$. This function we denote by $g(x) = \sqrt[n]{x} = x^{1/n}$. Next challenge is to show the continuity of $g(x)$. This is tricky.
Let $a > 0$ and we will show that $g(x)$ is continuous at $x = a$. Let $b = g(a) = a^{1/n}$ then $b > 0$ and $b^{n} = a$. Also we write $y = g(x) = x^{1/n}$ so that $y^{n} = x$. Now we can see that $$x - a = y^{n} - b^{n} = (y - b)(y^{n - 1} + y^{n - 2}b + \cdots + yb^{n - 2} + b^{n - 1})$$ Suppose that $x \to a^{+}$ so that $x > a$ and we need to show that $y = g(x) \to g(a) = b$. We can see clearly that $y > b$ and hence $$y^{n - 1} + y^{n - 2}b + \cdots + yb^{n - 2} + b^{n - 1} > nb^{n - 1}$$ It follows that $$0 < y - b < \frac{x - a}{nb^{n - 1}} $$ Now using sqeeze theorem and letting $x \to a^{+}$ we get $\lim_{x \to a^{+}}(y - b) = 0$ or $\lim_{x \to a^{+}}y = b$. Similarly we can handle $x \to a^{-}$.
The case $a = 0, b = 0$ requires a different approach. You should try that by using simple inequalities.
