OH! I came up a similar but much shorter proof:
Let $\epsilon\geq 0$, define $U_{\epsilon}:=\Big[\mathbb{E}(|X|^{p}|\mathcal{F})+\epsilon\Big]^{1/p}$ in $L_{p}(\Omega,\mathcal{F},\mathbb{P})$ and $V_{\epsilon}:=\Big[\mathbb{E}(|Y|^{q}|\mathcal{F})+\epsilon\Big]^{1/q}$ in $L_{q}(\Omega,\mathcal{F},\mathbb{P})$. Then, for each $\epsilon>0$, both $U_{\epsilon}$ and $V_{\epsilon}$ are uniformly bounded by $0$ from below.
Recall from the proof of regular Holder that we have, for all $x,y\geq 0$, that $$\dfrac{x^{p}}{p}+\dfrac{y^{q}}{q}-xy\geq 0,$$ for which we consider the first two derivatives in $x$ of the function on the LHS above.
This implies that for all $\omega$ and $\epsilon>0$, we have $$\Big|\dfrac{X(\omega)Y(\omega)}{U_{\epsilon}(\omega)V_{\epsilon}(\omega)}\Big|\leq\dfrac{1}{p}\Big|\dfrac{X(\omega)}{U_{\epsilon}(\omega)}\Big|^{p}+\dfrac{1}{q}\Big|\dfrac{Y(\omega)}{V_{\epsilon}(\omega)}\Big|^{q}.$$
Since both $1/U_{\epsilon}$ and $1/V_{\epsilon}$ are uniformly bounded, the expectation of both sides conditional upon $\mathcal{F}$ is well defined, and then it follows from the monotonicity that for almost every $\omega$, we have $$\dfrac{\mathbb{E}(|XY||\mathcal{F})}{U_{\epsilon}V_{\epsilon}}\leq\dfrac{1}{p}\dfrac{U_{0}^{p}}{U_{\epsilon}^{q}}+\dfrac{1}{q}\dfrac{V_{0}^{q}}{V_{\epsilon}^{q}}\leq\dfrac{1}{p}+\dfrac{1}{q}=1.$$
To finish, multiply both side by $U_{\epsilon}V_{\epsilon}$, and take $\epsilon\searrow 0$, then we have the desired inequality $$\mathbb{E}(|XY||\mathcal{F})\leq U_{0}V_{0}.$$
Recall Holder's Inequality for $(x) \in l^p$, $(y) \in l^q$ for conjugate $p,q$ is
$$\|xy\|_1 := \sum_{i=1}^\infty |x_i y_i| \leq \left( \sum_{i=1}^\infty |x_i|^p \right)^{1/p} \left( \sum_{i=1}^\infty |y_i|^q \right)^{1/q} =: \|x\|_p \|y\|_q.$$
So you're most of the way there by proving it for elements of unit norm. From here you just need a scaling argument.
Suppose $(x)$, $(y)$ are nonzero and as above. Define $(\tilde{x})$, $(\tilde{y})$ to be $(x/\|x\|_p)$, $(y/\|y\|_p)$ respectively. Then, by what you've done so far:
$$\sum_{i=1}^\infty |\tilde{x}_i \tilde{y}_i| \leq 1.$$
Then we get
$$
\require{cancel}
\begin{aligned}
\sum_{i=1}^\infty |x_i y_i| &= \sum_{i=1}^\infty |\|x\|_p\tilde{x}_i \|y\|_q \tilde{y}_i| \\
&= \|x\|_p\|y\|_q \sum_{i=1}^\infty |\tilde{x}_i \tilde{y}_i| \\
&\leq \|x\|_p\|y\|_q \cdot 1 \\
&= \|x\|_p\|y\|_q
\end{aligned}.$$
The case for when either $(x)$ or $(y)$ is $0$ should be trivial.
$\blacksquare$
Best Answer
We have $$ \mathbb{E}[|Z|^q] \leq \mathbb{E}\bigl[|X|^{q/p}|Y|\bigr] = \mathbb{E}\bigl[|X|^{q/p}\bigr] \mathbb{E}[|Y|] + \mathrm{cov}\bigl(|X|^{q/p},|Y|\bigr) \leq \mathbb{E}\bigl[|X|^{q/p}\bigr] \mathbb{E}[|Y|] $$ if $\mathrm{cov}\bigl(|X|^{q/p},|Y|\bigr) \leq 0$. This is the case for instance if $X$ and $Y$ are independent.
In general, the inequality is not true. Consider $X=Y=Z \sim \mathrm{Bernoulli}(1/2)$ and $p=q=2$. Then $$ \mathbb{E}[|Z|^2] = \frac12 > \frac14 = \mathbb{E}[|X|]\mathbb{E}[|Y|]. $$