Actually $\|u\|_{L^\infty(\eta B)}=\lim_{n\to\infty}\left(\int_{r_nB}|u|^n\right)^{\frac{1}{n}}=\lim_{n\to\infty}\|u\|_{L^n(r_nB)}.$
Denote the Lebesgue measure on $\mathbb{R}^n$ by $m$. The following fact is well known and even true in general measure space. Since it is easy to prove but I have no reference at hand, please let me write down a proof without citation.
Lemma: If $E\subset\mathbb{R}^n$ is measurable and $m(E)<\infty$, then for any measurable function $f$ defined on $E$, we have $\|f\|_{L^\infty(E)}=\lim_{p\to\infty}\|f\|_{L^p(E)}=\lim_{p\to\infty}\left(\int_E|f|^p\right)^{\frac{1}{p}}.$
Proo of Lemma: It suffices to consider the case $m(E)>0$ and $\|f\|_{L^\infty(E)}>0$. Given $0, there exists $E_c\subset E$, measurable and $m(E_c)>0$, such that $|f|\ge c$ on $E_c$. Then $m(E_c)^{\frac{1}{p}}\cdot c\le\left(\int_{E_c}|f|^p\right)^{\frac{1}{p}}\le \left(\int_E|f|^p\right)^{\frac{1}{p}}\le m(E)^{\frac{1}{p}} \|f\|_{L^\infty(E)}.$ Letting $p\to\infty$, we have $c\le\liminf_{p\to\infty}\|f\|_{L^p(E)}\le\limsup_{p\to\infty}\|f\|_{L^p(E)}\le\|f\|_{L^\infty(E)}.$ Since $c$ is arbitrary, the lemma follows.
Now for every $n\in\mathbb{N}$, there exists $N_n\in\mathbb{N}$, such that when $k\ge N_n$, $r_n\le r_k<\eta$, and hence $\|u\|_{L^k(r_nB)}\le\|u\|_{L^k(r_kB)}\le\|u\|_{L^k(\eta B)}.$ Letting $k\to\infty$, according to the lemma above, we have:
$\|u\|_{L^\infty(r_nB)}\le\liminf_{k\to\infty}\|u\|_{L^k(r_kB)}\le\limsup_{k\to\infty}\|u\|_{L^k(r_kB)}\le \|u\|_{L^\infty(\eta B)}.$ Since $\lim_{n\to\infty}r_n=\eta$, $\lim_{n\to\infty}r_n B=\eta B$, and hence $\lim_{n\to\infty}\|u\|_{L^\infty(r_nB)}=\|u\|_{L^\infty(\eta B)}$. The conclusion follows.