You are probably stuck because the random variable $N_n$ may assume far too many values for Kolmogorov's inequality to provide an effective upper bound. This suggests to deal separately with the case when $N_n$ is around $a_n$ (which, by Kolmogorov's inequality, should yield small values of $S_{N_n}-S_{a_n}$) and with the case when $N_n$ is far from $a_n$ (which, from the hypothesis that $N_n/a_n\to1$ in probability, should have small probability).
Hence, let us introduce, for a given positive $\varepsilon$, the event $$A_n=[(1-\varepsilon) a_n\leqslant N_n\leqslant (1+\varepsilon) a_n].$$
On the one hand, $N_n/a_n\to1$ in probability hence $A_n$ is typical in the sense that $\mathrm P(\Omega\setminus A_n)\to0$.
On the other hand, $|S_{N_n}-S_{a_n}|\leqslant |S_{N_n}-S_{(1-\varepsilon) a_n}|+|S_{a_n}-S_{(1-\varepsilon) a_n}|$ hence, on the event $A_n$,
$$
|S_{N_n}-S_{a_n}|\leqslant 2M_n,\qquad M_n=\sup\limits_{1\leqslant k\leqslant 2\varepsilon a_n}|T_k|,\qquad T_k=S_{(1-\varepsilon) a_n+k}-S_{(1-\varepsilon) a_n}.
$$
Now, we are back to the realm where Kolmogorov's inequality applies, and yields
$$
\mathrm P(M_n\geqslant x\sqrt{a_n})\leqslant (a_nx^2)^{-1}\mathrm{Var}(T_{2\varepsilon a_n})=(a_nx^2)^{-1}(2\varepsilon a_n)\sigma^2=2\varepsilon x^{-2}\sigma^2.
$$
Putting our two estimates together yields
$$
\mathrm P(|S_{N_n}-S_{a_n}|\geqslant 2x\sqrt{a_n})\leqslant\mathrm P(\Omega\setminus A_n)+\mathrm P(M_n\geqslant x\sqrt{a_n})\leqslant\mathrm P(\Omega\setminus A_n)+2\varepsilon x^{-2}\sigma^2.
$$
This proves that, for every positive $\varepsilon$,
$$
\limsup\limits_{n\to\infty}\ \mathrm P(|S_{N_n}-S_{a_n}|\geqslant 2x\sqrt{a_n})\leqslant2\varepsilon x^{-2}\sigma^2,
$$
hence $\mathrm P(|S_{N_n}-S_{a_n}|\geqslant2x\sqrt{a_n})\to0$ for every $x$, that is, $S_{N_n}/\sqrt{a_n}-S_{a_n}/\sqrt{a_n}\to0$ in probability.
By the usual central limit theorem, since $a_n\to+\infty$, $S_{a_n}/\sqrt{a_n}$ converges in distribution to a centered gaussian distribution with variance $\sigma^2$, hence $S_{N_n}/\sqrt{a_n}$ converges in distribution to the same centered gaussian distribution with variance $\sigma^2$.
This argument works, but in a sense it's overkill. You have a finite variance $\sigma^2$ for each observation, so $\operatorname{var}\left(\overline{X}_n\right)=\sigma^2/n$. Chebyshev's inequality tells you that
$$
\Pr\left(\left|\overline{X}_n - \mu\right|>\varepsilon\right) \le \frac{\sigma^2}{\varepsilon^2 n} \to 0\text{ as }n\to\infty.
$$
And Chebyshev's inequality follows quickly from Markov's inequality, which is quite easy to prove.
But the proof of the central limit theorem takes a lot more work than that.
Best Answer
I think there is confusion about the convergence statement for 'large' but finite $n$. Take, for example, the $X$ distribution $$p(x)=\frac{2}{\pi}(1+x^2)^{-2}$$ which has mean $0$ and variance $1$. The convergence statement is, for every $\epsilon>0$ there is an $n$ such that $\left|P(\bar S_n\leq z)-\Phi(z) \right|<\epsilon$ in some norm. This does not imply that the ratio $\frac{P(\bar S_n\leq z)}{\Phi(z)}$ is bounded for any finite $n$ for all $z$ on the real line. (I apologize, I know I did not define $\bar S_n=S_n/\sqrt{n}=\sqrt{n}\hat \mu$, Andrew's comment clarifies.)
Asymptotically, for the example distribution, up to constants (that depend on $n$, but not $z$), $P(\bar S_n\leq z)\sim \frac{1}{|z|^{3n}}$ and $\Phi(z)\sim \frac{e^{-z^2/2}}{|z|}$. These come from the leading terms in the Laurent series expansion of the CDF $\int_{-\infty}^z{p_X(t) dt}$ near $z=-\infty$. I am using the generous lower bound $(P(X_i\leq z\sqrt{n}))^n \leq P(\bar S_n\leq z)$. You can check these expressions here for the sample distribution and here for the normal approximation. So for every $n$, the ratio is unbounded as $z\to - \infty$.
Convergence implies that the interval in which such a bound holds can be made arbitrarily large. But unfortunately it is never the whole line for any finite $n$. The 'closeness' guaranteed by the CLT is as the norm of the difference, which does not translate to a bound in the ratio.