In the following blog post https://terrytao.wordpress.com/2008/06/18/the-strong-law-of-large-numbers/ one is presented with a nice account of the LLN.
Suppose that I have shown that if $(n_j)$ is a sequence of positive integers such that, for some $c>1$, we have $n_{j+1}/n_j>c=1+\delta$ then the means $\overline X_{n_j}$ converge almost surely to the expectation $\mathbb EX$. Note that if $m\leqslant n\leqslant m(1+\delta')$ then $$\tag{9} \overline X_m\leqslant \frac nm \overline X_n\leqslant (1+\delta')\overline X_n$$
which applies in particular to $(n_j)$ since $n_{j+k}>c^kn_j$. More precisely, the post asserts (which I have verified)
Suppose $X\geqslant 0$ and $(X_i)$ is a sequence of i.i.d. copies of $X$, with $\mathbb EX<\infty$. Assume the $(n_j)$ are lacunary (as just defined above). Then $\overline X_{n_j}$ converges to $X$ almost surely.
The post then says
Indeed, if we could prove the reduced version, then on applying that version to the lacunary sequence $n_j := \lfloor (1 + \varepsilon)^j\rfloor$ and using $(9)$ we would see that almost surely the empirical means $\overline{X}_n$ cannot deviate by more than a multiplicative error of $1+O(\varepsilon)$ from the mean $\Bbb E X$. Setting $\varepsilon := 1/m$ for $m=1,2,3,\ldots$ (and using the fact that a countable intersection of almost sure events remains almost sure) we obtain the full strong law.
I am failing to fill in the details here. Could someone do so?
Here is the argument from section 2.4 (page 65) of Probability: Theory and Examples (4th edition) by R. Durrett (freely available here).
Suppose that $n_j\leq m\leq n_{j+1}$. Since the $X$s are non-negative we get $S_{n_j}\leq S_m\leq S_{n_{j+1}}$ so that $${S_{n_j}\over n_{j+1}}\leq {S_m\over m}\leq {S_{n_{j+1}}\over n_j},$$ which implies $$\left({n_j\over n_{j+1}}\right){S_{n_j}\over n_{j}} \leq {S_m\over m}\leq \left({n_{j+1}\over n_j}\right) {S_{n_{j+1}}\over n_{j+1}},$$ and hence $$\left[\left({1\over 1+\varepsilon}\right)+o(j)\right]{S_{n_j}\over n_{j}} \leq {S_m\over m}\leq \left[\left(1+\varepsilon\right)+o(j)\right] {S_{n_{j+1}}\over n_{j+1}}.$$ Letting $j\to\infty$ and using your law of large numbers gives, almost surely,
$$\left({1\over 1+\varepsilon}\right) {\mathbb E}(X)\leq \liminf_m {S_m\over m} \leq \limsup_m{S_m\over m} \leq \left(1+\varepsilon\right) {\mathbb E}(X).$$
To be more specific, define
$$\Omega_\varepsilon=\left\{\omega\in \Omega: \left({1\over 1+\varepsilon}\right) {\mathbb E}(X)\leq \liminf_m {S_m(\omega)\over m} \leq \limsup_m{S_m(\omega)\over m} \leq \left(1+\varepsilon\right) {\mathbb E}(X)\right\},$$ so that the above argument gives $\mathbb{P}(\Omega_\varepsilon)=1$. Set $\Omega^*=\cap_{m=1}^\infty \Omega_{1/m}$ then $\mathbb{P}(\Omega^*)=1$, and for $\omega\in\Omega^*$ we have $S_m(\omega)/m\to\mathbb{E}(X)$.