Problem: Show that the conclusion is not true for $X=L^\infty[0,1]$.
Here is my attempt, which I now realize is incorrect:
Consider $f_n \in L^1[0,1] = Y$ defined by $f_n = n \cdot 1_{[0,1/n]}$. I already showed that no subsequence of $\{f_n\}$ converges weakly in $Y$. Let $T_n : X \to \Bbb{R}$ be defined by $T_n(f) = \int_X f \cdot f_n$. It's not hard to show $\{T_n\}$ is a bounded sequence. Hence, if Helly's theorem were true, there would exist a subsequence $\{T_{n_k}\}$ and $T \in X^*$ s.t. $T(f) = \lim_{k \to \infty} T_{n_k}(f)$. But $T(f) = \int_X f g$ for some $g \in Y$ and therefore $$\lim_{k \to \infty} \int_X f \cdot f_{n_k} = \int_X fg$$ for every $f \in X$. This means that $f_{n_k} \to g$ weakly in $Y$, which is a contradiction.
As you may have noticed, I wrongly thought that the Riesz Representation theorem holds for $L^\infty[0,1]$...such a nice, quick solution, too...Is there anyway of fixing it? I found this question asked elsewhere on MSE, but the proposed solution uses the Hahn-Banach theorem, which is currently not at my disposal.
I Know this answer comes much later than you probably would’ve liked but your argument does in fact hold. One may show that $T_n(g) := \int_{[0,1]} f_n g$ is a bounded linear functional for all $n \in \Bbb{N}$. The linearity comes from linearity of the integral. As for the bound, we see using Hölder's inequality that:
$$|T_n(g)|=\big|\int_{[0,1]} f_n \cdot g \space\big| \le \|f_n\|_1\|g\|_{\infty} ,\space \space g \in L^{\infty}[0,1] $$
$$ \Rightarrow \space \space \frac{T_n(g)}{\|g\|_{\infty}} \le \|f_n\|_1=1 \space\space\forall n \in \Bbb{N}$$ $$\Rightarrow \space \space \|T_n\|_* \le 1\space\space\forall n \in \Bbb{N}$$ and thus, $T_n$ is a bounded linear functional on $L^{\infty}[0,1]$ and your argument holds.