In the Math Stack Exchange post, I gave a proof based on Lemma 2 in Bai and Yin (1993). I will give an alternative proof.
Expressing $\sum_{1\leqslant i\neq i'\leqslant n}X_{i,j}X_{i',j}$ as
$\left(\sum_{i=1}^n X_{i,j}\right)^2-\sum_{i=1}^nX_{i,j}^2$ and considering dyadic numbers, we are reduced to show that
$$
\tag{1}\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{n}}\max_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum_{i=1}^\ell X_{i,j}\right\rvert\to 0\mbox{ almost surely}
$$
$$
\tag{2}\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} X_{i,j}^2\to 0\mbox{ almost surely}.
$$
By the Borel-Cantelli lemma, in order to prove (1), it suffices to show that
$$
\tag{1'}\forall\varepsilon>0, \sum_{n\geqslant 1}\mathbb P\left(\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{n}}\max_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum_{i=1}^\ell X_{i,j}\right\rvert>\varepsilon\right) <\infty$$
which reduces, by a union bound, to prove that
$$
\tag{1''}\forall\varepsilon>0, \sum_{n\geqslant 1}2^n\mathbb P\left( \frac{1}{2^{n}}\max_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum_{i=1}^\ell X_{i,0}\right\rvert>\varepsilon\right) <\infty$$
Let us prove (1''). Using Corollary 1.5 in 1 (applied with $r'=r=2$ and $B=\mathbb R$, hence $C_{r',B}=1$, we know that for each $q>0$, there exist constants $A(q)$ and $B(q)$ such that for each $x>0$ and each independent sequence $(Y_i)$,
$$
\mathbb P\left(\max_{1\leqslant \ell\leqslant N}\left\lvert \sum_{i=1}^\ell Y_i\right\rvert >x\right)\leqslant A(q)\int_0^1u^{q-1}\mathbb P\left(\max_{1\leqslant i\leqslant N}\lvert Y_i\rvert>xB(q)u\right)du+A(q)x^{-q}\left(\mathbb E\left[Y_i^2\right]\right)^{q/2}.
$$
Applying this inequality to $Y_i=X_{i,0}$, $x=2^n\varepsilon$ and $q=3$ shows that
$$
\sum_{n\geqslant 1}2^n\mathbb P\left( \frac{1}{2^{n}}\max_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum_{i=1}^\ell X_{i,0}\right\rvert>\varepsilon\right)\leqslant A(3)\sum_{n\geqslant 1}2^{2n}\int_0^1u^{2}\mathbb P\left( \lvert X_{i,0}\rvert>\varepsilon B(3)2^nu\right)du + A(3)\sum_{n\geqslant 1}2^{n}2^{-3n/2}\varepsilon^{-3/2}
$$
and finiteness of the first series follows from the elementary fact that $\sum_{n\geqslant 1}2^{2n}\mathbb P\left(Y>2^n\right)\leqslant 4\mathbb E\left[Y^2\right]$.
(2'') follows from an application of the similar argument as the usual strong law of large number, but the truncation level changes. Let
$Y_{i,j}=X_{i,j}^2$ and for a fixed $\varepsilon$ and $n$, let
$$
Z_{n,i,j}=Y_{i,j}\mathbf{1} \{Y_{i,j}\leqslant\varepsilon 2^{2n} \}-\mathbb E\left[Y_{i,j}\mathbf{1}\{Y_{i,j}\leqslant\varepsilon 2^{2n}\}\right],
$$
$$
W_{n,i,j}=Y_{i,j}\mathbf{1}\{Y_{i,j}>\varepsilon 2^{2n}\}-\mathbb E\left[Y_{i,j}\mathbf{1}\{Y_{i,j}>\varepsilon 2^{2n}\}\right],
$$
We thus have to show that
$$
\tag{2'}\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} Z_{n,i,j}\to 0\mbox{ almost surely}.
$$
$$
\tag{2''}\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} Z_{n,i,j}\to 0\mbox{ almost surely}.
$$
For (2'), it suffices to show (by the Borel-Cantelli lemma) that
$\sum_m\mathbb E\left[\left(\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} Z_{n,i,j}\right)^2\right]<\infty$.
For (2''), note that $$ \left(\{\max_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum_{i=1}^{2^n} Z_{n,i,j}\neq 0 \}\right)\subset\bigcup_{i,j=1}^{2^n}\left(\{Y_{i,j}>\varepsilon 2^{2n}\}\right).$$
1 Deviation inequalities for Banach space valued martingales differences sequences and random fields Davide Giraudo ESAIM: PS 23 922-946 (2019)
Best Answer
Of course, this is not true. E.g., suppose that the $a_n$'s are independent random variables each uniformly distributed on the finite set $B$, of cardinality $|B|\ge1$. (With $\nu_l(b)$ and $\nu_u(b)$ completely unspecified, the condition $\nu_l(b) \le P[(a_n = b|a_1,\ldots,a_{n-1}) \le \nu_u(b)$ is not a restriction at all. In our present case, we can take $\nu_l(b)=\nu_u(b)=P(a_n=n)=1/|B|$ for all $b\in B$.)
Then for any real $T_b$ and all $n>T_b^2|B|^2$ $$E\Big(\frac1n\,\sum_{k=1}^n 1(a_k = b) \Big) =\frac1n\,\sum_{k=1}^n E1(a_k = b)=\frac1{|B|} \not\le \frac{T_b}{\sqrt{n}}.$$
Added in response to a comment by the OP: In general, $$E\Big(\frac1n\,\sum_{k=1}^n 1(a_k = b) \Big) =\frac1n\,\sum_{k=1}^n E1(a_k = b) \\ =\frac1n\,\sum_{k=1}^n P(a_k = b) =\frac1n\,\sum_{k=1}^n EP(a_k = b|a_1,\ldots,a_{n-1}) \\ \le\frac1n\,\sum_{k=1}^n E\nu_u(b) =\nu_u(b).$$ Similarly, $$E\Big(\frac1n\,\sum_{k=1}^n 1(a_k = b) \Big) \ge\nu_l(b).$$