What i'm proving is in the title. Essentially, I know that it holds for rational numbers. I want to prove it for all real numbers. The following is the definition for $a^x$ that I'll be using.
Let $\{r_n\}$ be any sequence of rationals that converge to $x$. Then, the exponential of the base $a$ is defined:
$$a^x = \lim_{n \to \infty} a^{r_n}$$
Proof Attempt:
Let $\{r_n\}$ be any sequence of rational numbers that converges to $x$.
This sequence is convergent, so it is bounded. By the Bolzano-Weierstrass Theorem, this has a convergent subsequence. We denote this subsequence by $\{s_n\}$.
This subsequence converges to $x$ as well. If it didn't, then there would exist an $\epsilon > 0$ such that for all $N>0$, $n>N$ and $|s_n-x| \geq \epsilon$. But since this subsequence is a part of the original sequence, which does converge to $x$, it follows that such an $\epsilon$ cannot exist.
If the subsequence is decreasing, then:
$$\forall n \in \mathbb{N}: s_n > 0$$
So, it follows that:
$$\forall n \in \mathbb{N}: a^{s_n} > 1$$
In the limit, it would follow that $a^x > 1$.
If the subsequence is increasing, then:
$$\exists N > 0: n > N \implies s_n > 0$$
$$\implies \exists N>0: n > N \implies a^{s_n} > 1$$
In the limit, it follows that $a^x > 1$. This proves the desired result.
Does the proof above work? If it doesn't, why? How can I fix it?
It doesn't work, because you merely proved that $a^{s_n}\ge1$, if we assume the map is continuous.
To fix it, we'll assume the following exponentiation fact:
Consider any sequence $r_n$ that converges to $x$. Then we know that for $\varepsilon=\frac x2$, eventually every term in the sequence $a^{r_n}$ is greater than $1+\kappa$ for $\kappa=a^{\frac x2}-1>0$.
Thus, since eventually every term is greater than $1+\kappa$, the limit is at least $1+\kappa$ (note that it could be equal to $1+\kappa$, but this is still $>1$).
Of course, if you haven't already done so, you also need to show that this limit exists and is well-defined (independent of which sequence you choose).