How to show that regularly varying distributions are heavy tailed? Let distribution $F$ be regularly varying with index $\alpha \geq 0$ (denote $F \in R_{\alpha}$), i.e. its tail $\bar{F} = 1 - F$ satisfies
$\lim_{x\rightarrow\infty} \frac{\bar{F}(xy)}{\bar{F}(x)} = y^{-\alpha}$ for any $y>0$.
i) Show that $\bar{F}(x) = x^{-\alpha}L(x)$, where $L(x)$ is slowly varying function.
ii) Show that $F$ is heavy-tailed.

My attempt for the first part was by using a definition of slowly varying function, i.e. $L(x)$ is slowly varying if $\lim_{x\rightarrow\infty}\frac{L(xy)}{L(x)} =1$ for any $y>0$. I simply plugged the $\bar{F}(x)$ equation in the i) problem and got equality. Is this a correct way to show it? For the second part, I have no idea how to show that if $F$ is regularly varying distribution, then it is heavy-tailed.

This problem requires special care because no assumption of continuity of $F$ has been made.  It is possible that the support of this distribution omits arbitrarily long positive intervals of numbers or is singular (point-like).  This precludes more elementary solutions based on analyzing the derivative of $F,$ which needn't exist.
The additional difficulty in (ii), which is your main question, arises because you have too much information: "slowly varying" is not necessary.  All that we need for showing $F$ is heavy tailed is that $L(x)$ eventually (that is, for large $x$) stays away from zero and infinity--it doesn't have to converge to anything.  However, this fact is an immediate implication of the limit definition you helpfully supplied, so let's begin there.
Suppose then (as proven by part (i) of the question) there exist $\upsilon \gt \lambda \gt 0$ and $x_0$ for which $\lambda L(x) \le L(x) \le  \upsilon L(x)$ for sufficiently large $x;$ that is,
$$\lambda x^{\alpha}\bar F(x) \le x^{\alpha}\bar F(x) \le  \upsilon x^{\alpha}\bar F(x)\tag{*}$$
for all $x\ge x_0.$
What does "heavy-tailed" mean?  At https://stats.stackexchange.com/a/168918/919 I quote a standard definition:

$F$ is heavy-tailed when $\int_{\mathbb{R}} e^{tx} \mathrm{d}F(x) = \infty$ for all $t\gt 0.$

Intuitively this says $\bar F$ cannot decrease exponentially. The key part of the integral is over the positive real numbers, because this is where $e^{tx}$ "blows up" exponentially for $t\gt 0$ and $x$ large.  (When $x$ is negative, $e^{tx}$ gets really small, thereby creating no problems with the integral.)
The idea behind the following solution is to characterize "exponential rate" as "any rate faster than $x\to x^{\alpha-1}.$"  Given $t\gt 0$ and $\alpha \gt 0,$ eventually (for sufficiently large $x$), apply this insight to conclude $e^{tx}$ will exceed $(tx)^{\alpha-1}.$  We can prove it by setting $k = \lceil\alpha\rceil$ and underestimating the exponential with the $k^{\text{th}}$ term in its power series.  For all $x \ge k!/t$ this gives
$$e^{tx} \gt \frac{(tx)^k}{k!} \ge \frac{tx}{k!}(tx)^{k-1} \ge (tx)^{k-1} \ge (tx)^{\alpha-1}.\tag{**}$$
If necessary, increase $x_0$ to $k!/t$ so that all the inequalities in $(*)$ and $(**)$ hold.
Let us, therefore, (1) underestimate the integral by integrating over large $x$ only; (2) apply the preceding inequality; and (3) write it in terms of $\bar F(x) = 1 - F(x),$ for which $\mathrm{d}F = -\mathrm{d}\bar F.$  To accomplish $(3)$ we integrate by parts:
$$\begin{aligned}
\int_\mathbb{R} e^{t x} \mathrm{d}F(x) &\ge\int_\infty^{x_0}e^{t x}\mathrm{d}\bar F(x) \\
&\ge \int_\infty^{x_0} (tx)^{\alpha-1}\mathrm{d}\bar F(x) \\
&= \lim_{N\to\infty}\left[ (tx)^{\alpha-1} \bar F(x)\bigg|_N^{x_0} + t\int_{x_0}^N (tx)^{\alpha-1}\bar{F}(x)\mathrm{d}x\right].
\end{aligned}\tag{***}$$
Care is needed here because both of the limiting terms may diverge, but in opposite directions, and thereby cancel: thus, we cannot just thoughtlessly split this into two separate limits.  (This is precisely the difficulty that would arise if we did not underestimate $e^{tx}$ with the power $(tx)^{\alpha-1}.$)  Instead, consider each in turn.  First,
$$(tx)^{\alpha-1} \bar F(x)\bigg|_N^{x_0} = t^{\alpha-1}\left(N^{\alpha-1} \bar F(N) - x_0^{\alpha-1} \bar F(x_0)\right).$$
Because, once $N$ is large enough, $\bar F(N)$ is bounded between two multiples of $N^{-\alpha},$ the product $N^{\alpha-1}\bar F(N)$ is bounded between two multiples of $N^{-1},$ which converges to zero.  This leaves us a similar-looking expression which, by the same analysis, is bounded (for sufficiently large $x_0$) between multiples of $x_0^{-1}.$  Consequently, the entire expression can be made arbitrarily small when $x_0$ is sufficiently large (and $N\ge x_0,$ of course).
In the second term, the lower bound can be evaluated easily:
$$t\int_{x_0}^N (tx)^{\alpha-1}\bar{F}(x)\mathrm{d}x \ge t\int_{x_0}^N (tx)^{\alpha-1}\lambda x^{-\alpha}\mathrm{d}x = \lambda t^{\alpha}\int_{x_0}^N x^{-1}\mathrm{d}x =  \lambda t^{\alpha}\left(\log N - \log x_0\right).$$
No matter what value $x_0$ might have, the positivity of $t$ and $\lambda$ imply this term diverges to $+\infty$ as $N$ grows large.  Consequently, the limit in $(***)$ consists of one term approaching zero plus one term that diverges, demonstrating that $F$ is heavy-tailed, QED.

One interesting implication of this approach is that any distribution that has an infinite absolute moment (of degree greater than $0$) of the positive part of that distribution is heavy-tailed.  This result is immediate.
The converse is not true, though: there are heavy-tailed distributions for which all absolute moments are finite.  One of them is determined by the survival function expressed as
$$\bar F(x) = \exp\left(-\frac{x}{\log x} + e\right)$$
for $x \ge e = \exp(1).$  This almost, but not quite, decays exponentially.
By comparing $\log x$ to $\sqrt x$ (which exceeds any given multiple of $\log x$ for large $x$) you can readily estimate the moments as well as the integral involved in the definition of heavy tails.  The integral of $e^{tx}\mathrm{d}\bar F(x)$ diverges (evaluate it on the interval $x \ge \exp(2/t)$ for instance) but the moment of order $k\gt 0$ is bounded above by $2e\Gamma(2k+1).$
