If $fg\in L_1$ for every $f\in L_p$, show that $g\in L_q$

Let us assume that $\mu$ is a finite measure, to simplify exposition. See the bottom of this post for the general, $\sigma$-finite case.

To prove that $f\mapsto \int fg$ is a bounded linear functional on $L^p$, a standard trick is the use of the uniform boundedness principle. Indeed, let $$ g_n(x):=\begin{cases} n, & \lvert g(x) \rvert \ge n, \\ g(x), & \lvert g(x)\rvert <n, \end{cases}$$ (this is known as a truncation of $g$). Define a linear functional on $L^p$ as $$ T_n f:=\int g_n f\, d\mu.$$ This is a bounded functional, because the Hölder inequality gives $$ \lvert T_n f\rvert \le \lVert g_n\rVert_{p'}\lVert f\rVert_p, $$ and $\lVert g_n\rVert_{p'}<\infty$ since $g_n$ is bounded and $\mu$ is finite.

Now, define another functional as $$Tf:=\int gf\, d\mu.$$ So far, we do not know whether $T$ is bounded or not. However, we know that, by dominated convergence, for each $f\in L^p$ we have $T_n f\to Tf$. Moreover, using that $$ \lvert g_n(x)\rvert \le \lvert g(x)\rvert, $$ we can estimate $$ \lvert T_n f\rvert \le \int \lvert gf\rvert\, d\mu <\infty.$$ This last inequality, and the fact that each $T_n$ is a bounded functional, imply by the uniform boundedness principle that there exists $C>0$ such that $$ \lVert T_n\rVert_{(L^p)^\ast}\le C, \quad \text{ for all }n,$$ where, as usual, $\lVert T_n\rVert_{(L^p)^\ast}:=\sup\{ \lvert T_n h\rvert\ |\ h\in L^p, \|h\|_p=1\}$. And since $T_nf\to Tf$, this implies that $\lVert T\rVert_{(L^p)^\ast}\le C$, which is what you wanted to prove.

The $\sigma$-finite case is essentially the same, with a small additional technical detail. Notice that we used finiteness only to ensure that $g_n\in L^{p'}$, given that $g_n$ is bounded; this is not true if $\mu$ is not a finite measure. To circumvent this, let $\Omega_n$ be a sequence of sets such that $\mu(\Omega_n)<\infty$ and $\bigcup \Omega_n =\Omega$. Redefine $g_n$ so that it reads $$ g_n(x):=\begin{cases} n, & \lvert g(x) \rvert \ge n\ \text{and }x\in \Omega_n, \\ g(x), & \lvert g(x)\rvert <n\ \text{and }x\in \Omega_n,\\ 0,& x\notin \Omega_n. \end{cases}$$ Now, $g_n$ is bounded and supported in a set of finite measure. Thus, $g_n\in L^{p'}$.

The proof goes on from this point exactly as before.