Suppose $L^q(X,\mu)$ is separable (i.e. admits a countable dense subset). I wish to prove that every sequence $\{f_n\}$ in $L^p$ that satisfies $\sup_n \|f_n\|_p < \infty$ has a weakly convergent subsequence, i.e. a subsequence $f_{n_k}$ and $f \in L^p(X,\mu)$ such that for every $g \in L^q(X,\mu)$ we have
$$\int f_{n_k}g \to \int fg.$$
My attempt: Let $\{x_n\}$ be a countable dense subset of $L^q(X,\mu)$. By Holder's inequality we know that $\int f_nx_1$ is a bounded sequence of real numbers and hence we can find a subsequence $f_n^1$ such that $\int f_n^1 x_1$ converges to a real number $a_1$. Now again we find that $ \int f_n^1 x_2 $ is a bounded sequence of real numbers and so there is a subsequence $ \{f_n^2\} \subseteq \{f_n^1\} $ for which $\int f_n^2 x_2$ converges to a real number $a_2$. Continuing this process and taking the diagonal $f_k^k$ we see that given any $x_n$, $$\lim_{k \to \infty} \int f_k^k x_n= a_n.$$
Also, we may get a linear functional on all of $L^q(X,\mu)$ by setting $$l(x) = \lim_{k \to \infty} \int f_k^k x.$$
My question is: Why is this well-defined, namely why does $\lim_{k \to \infty} \int f_k^k x$ exist? If we know it does then the UBP gives that $l$ is bounded and so the weak limit $f$ is furnished from the proof that the dual of $L^q$ is $L^p$.
Assume first that $x_n$ is a Cauchy sequence in $L^q$. Note that $$\left|\int f_k^kx_n-\int f_k^kx_m\right|\leq\int|f_k^k||x_n-x_m|\leq \|f_k^k\|_p\|x_n-x_m\|_q\tag{1}$$
Because $f_k^k$ is bounded in $L^p$, we conclude that the convergence in $(1)$ is uniformly in $k$ (so we can change the order of limits). Moreover, we also conclude from $(1)$ that $\lim_{n\to\infty} a_n=a$ for some $a\in\mathbb{R}$. Therefore $$\lim_{k\to\infty}\lim_{n\to\infty}\int f_k^k x_n=\lim_{n\to\infty}\lim_{k\to\infty}\int f_k^k x_n=\lim_{n\to\infty }a_n=a$$