How to proof the existence of the nth-roots using the Intermediate Value Theorem Given $y>0$ prove that $\exists\, x>0$ such that $x^n$ = $y$.
We have to find $b>0$ such that $b^n$ > $y$.
We have 3 cases:
1) $y>1$
We have that $y^n>y$, so we can set $b=y$.
2) $y<1$ 
Let's set $b=1$.
3) $y=1$
This is simply $x=1$.
The function $f(x) = x^n$, $f:[0,b] \rightarrow ℝ$ accomplishes the Intermediate Value Theorem, i.e, $\exists\, x \in (0,b)$ such that $f(x)= y$.
Is that right?
 A: Your argument is more or less correct.
If $a = 0$ then $a^n = 0 < y$.
If $y < 1$ and if $b = 1$ then $b^n = 1 > y$
If $y > 1$ and if $b = y$ then $b^n = y^n > y$.
If $y = 1$ then .... well we could simply do $b = 2$ and $2^n > 1 = y$... or we could simply point out $1^n = y$ so that is the $n$-th root we want.
So any event you have $a^n < y < b^n$ so by intermediate value theorem there is a $c \in [a,b]$ so that $c^n = y$.
...
However that makes me very uneasy because we must assume $f(x) = x^n$ is a continuous function.
A: It appears safe to assume $1 \le n \in \Bbb N$, the natural numbers.
If we grant that $f(x) = x^n$ is in fact continuous, then it is but a short step, via the Intermediate Value Theorem, to see that $n$-th roots must exist:
Let $y > 0$.  Clearly, 
$f(0) = 0^n = 0 < y; \tag 1$
also,
$f(1 + y) = (1 + y)^n = 1^n + n \cdot 1^{n - 1} y + \dfrac{n!}{2!(n - 2)!} 1^{n -2} y^2 + \ldots + y^n$
$= 1 + n y + \ldots + y^n = y + 1 + (n - 1)y + \ldots + y^n > y + c, \tag 2$
with $c = 1 + (n - 1)y + \ldots + y^n > 0$; hence
$f(1 + y) > y; \tag 3$
we thus have
$0 = f(0) < y < f(1 + y), \tag 4$
and it now follows from the IVT that there is an $x \in (0, 1 + y)$ with
$y = f(x) = x^n, \tag 5$
and we are done!  
Done, that is, except for need to address the continuity of $f(x) = x^n$.  I would ordinarily feel it is safe to assume this as well, but since fleablood expressed concern that this fact be demonstrated, I offer the following:
If one accepts the use of a little elementary calculus, i.e. that
$f'(x) = nx^{n - 1}, \tag 6$
then since continuity follows from differentiability, then we may rest here.  However, the logical question as to how we know $f(x)$ is differentiable and that (6) binds lingers on; of course, an $\epsilon$-$\delta$ argument is the most obvious way to address this issue.  One can also side with Rob Arthan, who mentioned in his comment to the question itself that the continuity of $f(x) = x^n$ follows from the continuity of a product of continuous functions; but guess what?  But that assertion also typically requires an $\epsilon$-$\delta$ argument; thus all we attain by invoking it without proof is sweeping $\epsilon$-$\delta$ under the rug, as it were.  So I guess what it boils down to is we need to present the $\epsilon$-$\delta$ proof, or shut up and go home.  Therefore in favor of the former I flesh out the details of the argument I suggested in a comment to the main post:  we use the well-known factorization
$x_2^n - x_1^n = (x_2 - x_1) \left ( \displaystyle \sum_{k = 0}^{n - 1} x_2^{n - 1 - k}x_1^k \right ), \tag 7$
which implies
$\vert f(x_2) - f(x_1) \vert = \vert x_2^n - x_1^n \vert  = \vert x_2 - x_1 \vert  \left \vert \displaystyle \sum_{k = 0}^{n - 1} x_2^{n - 1 - k}x_1^k \right \vert; \tag 8$
under the assumption that $x_1, x_2 \ge 0$, which is appropriate since we are addressing the case $y > 0$, and hence only need consider $f(x) = x^n$ on $[0, \infty)$, we have
$\left \vert \displaystyle \sum_{k = 0}^{n - 1} x_2^{n - 1 - k}x_1^k \right \vert = \displaystyle \sum_{k = 0}^{n - 1} x_2^{n - 1 - k}x_1^k; \tag 9$
we now fix $x_1 \in [0, \infty)$ and choose $0 < M \in \Bbb R$ such that 
$x_1 < M; \tag {10}$
then if 
$0 \le x_2 < M\tag{11}$
as well, we see that
$\displaystyle \sum_{k = 0}^{n - 1} x_2^{n - 1 - k}x_1^k < \sum_{k = 0}^{n - 1} M^{n - 1} = nM^{n - 1}; \tag{12}$
(8) thus yields
$\vert f(x_2) - f(x_1) \vert = \vert x_2^n - x_1^n \vert  \le nM^{n - 1}\vert x_2 - x_1 \vert  ; \tag{13}$
if we now choose any $\epsilon > 0$ and take
$\delta < \min \left \{ \dfrac{\epsilon}{nM^{n - 1}}, M - x_1 \right \}, \tag{14}$
then with
$\vert x_2 - x_1 \vert < \delta \tag{15}$
we have
$-\delta < x_2 - x_1 < \delta, \tag{16}$
or
$x_1 - \delta < x_2 < x_1 + \delta < x_1 + (M - x_1) = M; \tag{17}$
since now $x_1, x_2 < M$ we may apply (13) in concert with (14):
$\vert f(x_2) - f(x_1) \vert < nM^{n - 1} \vert x_2 - x_1 \vert < nM^{n - 1} \dfrac{\epsilon}{nM^{n - 1}} = \epsilon, \tag{18}$
which shows that $f(x)$ is continuous at $x_1$; since $x_1 \in \Bbb R$ is arbitrary, we see that $f(x) = x^n$ is indeed continuous on $[0, \infty)$.
