Let $\psi \in C_c(\mathbb{R}^n)$ s.t. $\text{supp}\; \psi \subseteq B_1(0)$ and $\|\psi\|_{L^p(\mathbb{R}^n)} = 1$. For $k \in \mathbb{N}$ define $f_k: B_1(0) \to \mathbb{R}$ by $f_k(x) = k^{n/p}\psi(kx)$. Then $f_k(x) \to 0$ for all $x \neq 0$, since for large enough $k$ we get $kx \notin B_1(0)$ hence $\psi(kx) = 0.$
I wanna show that $f_k$ weakly converges to $0$, i.e. that $T(f_k) \to 0$ for all $T \in L^p(B_1(0))^*.$ Since $B_1(0)$ (with the lebesgue measure) is $\sigma$-finite, we know that for every linear functinoal $T$ in $L^p(B_1(0))$ there exists some $g \in L^q(B_1(0))$ (where $q$ is the dual exponent of $p$) such that
$$T(f_k) = \int_{B_1(0)}gf_k \;\mathrm{d\lambda^n}.$$ Now I'd like to use dominated convergence to show that this integral convergences to $0$ whence we're done. But I have trouble coming up with an upper bound for $gf_k$.
Here is a different idea: assume $p>1$. Let me prove that $\|f\|_{L^1} \to0$: $$ \int_{B_1} |f_k| = \int_{B_{1/k}} |f_k| = \int_{B_{1/k}} k^{n/p} |\psi(kx)|dx = k^{\frac np -n} \int_{B_1} |\psi(x)|dx \to 0. $$ Since $\|f_k\|_{L^p(B_1)} = \|\psi\|_{L^p}$, we have that $f_k$ converges strongly to zero in $L^q(B_1)$ for all $1\le q<p$. Hence weakly to zero in $L^p(B_1)$.