Proving that $\lim_{n \to \infty} x^{1/n} = 1$ at $x \gt 0$

Its required to prove that $|x^{1/n} -1| \lt \epsilon$ for $\epsilon \gt 0$ and $n \ge N$ where $N \in \mathbb N$.
Let $x^{1/n} -1 = h$ for $h \gt 0$. So $x = (1+h)^{n}$, and we now need to prove that $h \lt \epsilon$. The binomial theorem gives $x \gt nh $ so $h \lt \frac{x}{n}$. Now its sufficient to choose $n $ such that $\frac{x}{n} \lt \epsilon$, i.e. $n \gt\frac{x}{\epsilon}$. So the given condition is satisfied for all $ N \ge [\frac{x}{\epsilon}] + 1.$(The square brackets stand for the floor function.)
Is this in form and content correct?

It's almost correct :) Like already mentioned your proof only works for $x\geq 1$ because for $0\leq x<1$ we get some negative terms in the expansion of $(1+h)^n$ since $h<0$.

But it's easy to fix it. You can show by induction that $(1+h) ^n\geq 1+hn$ even for $h\geq -1$. See here https://en.m.wikipedia.org/wiki/Bernoulli%27s_inequality Hence, we get $$x-1=(1+h) ^n-1\geq hn. $$ After that your argumentation works.

Let $x^{(1/n)} - 1 = h$ for $h>0.~~$ So $x=(1+h)^{(1/n)}$.

No, this is wrong. The intermediate conclusion is that
$x = (1 + h)^n$.

An easier approach, if permitted, is to let $y = \log(x)$.

Then, regardless of whether $y$ is positive, negative, or zero, you have that

$$\lim_{n\to\infty} \frac{y}{n} = 0.$$

Therefore, it is routine to set up an $\varepsilon$ demonstration, choosing $N$ large enough so that for all $n \geq N$, you have that $\displaystyle e^{(y/n)}$ is in the neighborhood of $(1 - \varepsilon, 1 + \varepsilon).$

That is, you want to choose $N$ large enough so that

$\displaystyle \left|\frac{y}{N}\right| < \log(1 + \varepsilon)$

and

$\displaystyle \left|\frac{y}{N}\right| < \left|\log(1 - \varepsilon)\right|.$

Note
You can assume, without loss of generality, that $\varepsilon < 1$. That is, if the demonstration holds for a specific $\varepsilon < 1$, then the demonstration holds for any larger $\varepsilon_1$.

This is because
$0 < \varepsilon < \varepsilon_1 \implies (1 - \varepsilon, 1 + \varepsilon) \subseteq (1 - \varepsilon_1, 1 + \varepsilon_1).$