Consider a strictly increasing sequence $q_0<q_n<q_{n+1}<q<d$ such that $q_n\to q$ as $n\to \infty$ with $d>q$. Let $B\subset \Bbb R^d$ be a ball, so that $L^{q}(B)\subset L^{q_{n+1}}(B)\subset L^{q_{n}}(B)\subset L^{q_0}(B)$. Consider a sequence $(u_n)_n\subset L^q(B)$ such that $\|u_n\|_{L^{q_n}(B)}=1$ and $u_n\to u$ in $L^{q_0}(B)$.
I want to prove or disprove that $\|u\|_{L^{q}(B)}=1$. Note that by Fatou's lemma, we get $$\|u\|_{L^{q}(B)}\leq \liminf_{n\to\infty}\|u_n\|_{L^{q_n}(B)}=1.$$
The result is false.
Here is a detailed counter-example. Let $d=2$ and let $B \subset \Bbb R^2$ be any ball containing $[0,1]^2$.
For each $n \in \Bbb N$, let $q_n = \frac{3}{2} - \frac{1}{2(n+1)}=\frac{3n+2}{2n+2}$. So we have $1=q_0<q_n<q_{n+1}<q=\frac{3}{2}<2=d$ and $q_n\to q$ as $n\to \infty$.
Now, for each $n \in \Bbb N$, let us define $u_n : B \rightarrow \Bbb R$ by $$u_n(x,y)=(n+1)^\frac{1}{q_n}\chi_{\left [0,\frac{1}{n+1} \right]}(x)\chi_{[0,1]}(y)=(n+1)^\frac{2n+2}{3n+2}\chi_{\left [0,\frac{1}{n+1} \right]}(x)\chi_{[0,1]}(y)$$ It follows immediately that $$ \int_B |u_n(x,y)|^{q_n} dxdy = \int_B (n+1)\chi_{\left [0,\frac{1}{n+1} \right]}(x)\chi_{[0,1]}(y) dxdy =1$$ So, for all $n \in \Bbb N$, $\|u_n\|_{L^{q_n}(B)}=1$.
Now, note that $u_n\to 0$ in $L^{q_0}(B)$. In fact, $q_0=1$ and we have $$ \|u_n\|_{L^{q_0}(B)} = \int_B |u_n(x,y)| dxdy = \int_B (n+1)^\frac{2n+2}{3n+2}\chi_{\left [0,\frac{1}{n+1} \right]}(x)\chi_{[0,1]}(y)= (n+1)^{\frac{2n+2}{3n+2} -1}= \frac{1}{n^\frac{n}{3n+2}}$$ So $\lim_{n \to \infty} \|u_n\|_{L^{q_0}(B)} = 0$. So $u_n\to 0$ in $L^{q_0}(B)$. But $\|0\|_{L^{q}(B)}=0 \neq 1$.