You should proceed with the definition $a^{x} = \lim_{n \to \infty}a^{x_{n}}$ where $x_{n}$ is a sequence of rationals tending to $x$ and $a > 0$. This route is bit difficult compared to the standard route of using logarithms via integral. But providing a rigorous justification of all the usual properties of exponents using the above definition turns out to be a good exercise in real analysis.
The first thing one needs to do is to show that the limit $\lim_{n \to \infty}a^{x_{n}}$ exists and the definition is unambiguous i.e. if $x_{n}, y_{n}$ are sequences of rationals both tending to $x$ then $\lim_{n \to \infty}a^{x_{n}} = \lim_{n \to \infty}a^{y_{n}}$. The algebraic properties like $a^{x + y} = a^{x}a^{y}$ don't pose a major challenge.
About your inequalities let us say we have $a > b > 0$ and $x_{n}$ is a sequence of rationals tending to $x$. We need to show $a^{x} > b^{x}$. Clearly by the rules of rational exponents we have $a^{x_{n}} > b^{x_{n}}$ but taking limits as $n \to \infty$ weakens the inequality to $\geq $. So what we need are the following inequalities $$ra^{r - 1}(a - b) > a^{r} - b^{r} > rb^{r - 1}(a - b)$$ and $$sa^{s - 1}(a - b) < a^{s} - b^{s} < sb^{s - 1}(a - b)$$ where $r, s$ are rationals with $0 < s < 1 < r$. These are established in this answer.
Let $x > 1$ then we can suppose after a certain value of $n$, $x_{n} > 1$. Clearly then we have $$a^{x_{n}} - b^{x_{n}} > x_{n}b^{x_{n} - 1}(a - b)$$ Taking limits as $n \to \infty$ we get $$a^{x} - b^{x} \geq xb^{x - 1}(a - b) > 0$$ so that $a^{x} > b^{x}$. Similarly we can use other inequality (dealing with $s$) to handle the case when $0 < x < 1$. For negative values of $x$ the inequality is reversed i.e. $a^{x} < b^{x}$. This can be easily done via the rule that $a^{-x} = 1/a^{x}$.
