Showing that $q(f) \leq ||f||$ for weak $L^p(\mathbb{R}^n)$ norm $||.||$ and quasi-norm $q(.)$ with $f$ belonging to weak $L^p(\mathbb{R}^n)$ Let $f \in L^p(\mathbb{R}^n)$ for $p \in (1, \infty)$, take $\mu$ to be the Lebesgue measure and define
$$q(f) = \sup_{c > 0}c\mu\left(\{x \in \mathbb{R}^n: |f(x)| > c\}\right)^{1/p}$$
$$\mathcal{C} = \{S \subset \mathbb{R}^n: \text{S is measurable}, \mu(S) \in (0, \infty)\}$$
that is $\mathcal{C}$ consists of all measurable subsets of $\mathbb{R}^n$ with a positive finite measure. (To me the LaTeX is not rendering the set symbols, hence a verbal description).
$$||f|| = \sup_{S\in \mathcal{C}}\mu(S)^{(1-p)/p}\int_{S}|f|dx$$
I am trying to show that $q(f) \leq ||f||$ but I have trouble making a connection between the measurable sets $S\in \mathcal{C}$ and the parameter $c > 0$. Namely, on the one hand,
1.) $$q(f) \equiv \sup_{c>0}\mu\left(\{x \in \mathbb{R}^n: |f(c)| > c\}\right)^{1/p}\leq \sup_{c>0}c\left(\frac{1}{c}\int_{R(c)}|f|dx\right)^{1/p}$$
where $R(c) = \{x \in \mathbb{R}^n: |f(x)| > c\}$.
And on the other hand, if $c > 0$, then
2.) $$||f|| = \sup_{S\in \mathcal{C}}\mu(S)^{(1-p)/p}\int_S|f|dx \geq |S(c)|^{(1-p)/p}\int_{S(c)}|f|dx \Longleftrightarrow$$
$$||f|| \geq \mu\left(S(c)\right)^{(1-p)/p}\int_{S(c)}|f|dx \geq \mu\left(S(c)\right)^{(1-p)/p}c\mu\left(\{x \in S(c): |f(x)| > c\}\right)$$
(Main point:) That is, I have to somehow either show that for $c > 0$, $\frac{c}{c^{1/p}}$ is related to a the measure of any element of $\mathcal{C}$ raised to $\frac{1-p}{p}$, and massage the $\frac{1}{p}$ out of the 1.) integral by bounding the expression above, or
I have to argue why the second expression can somehow be bounded above when $S(c)$ is switched to $\mathbb{R}^n$.
Unfortunately I do not have the faintest of idea how to approach the problem from either directions.
 A: Notation
First we laid down some definitions and notation. Throughout this posting,  $(X,\mathscr{B},\mu)$ is a measure space.

Definition: $f\in L^{\infty, p}(\mu)$ if there exists $A>0$ such that $\mu(|f|>\lambda)\leq \frac{A^p}{\lambda^p}$ for all $\lambda>0$.

It follows from Markov-Chebyshev's inequality that $L^p(\mu)\subset L^{\infty,p}$: $\mu(|f|>\lambda)\leq\frac{\|f\|^p_p}{\lambda^p}$.

Lemma 1: If $f\in L^{\infty, p}(\mu)$
\begin{align}
[f]_p&:=\inf\{A: \lambda\big(\mu(|f|>\lambda)\big)^{1/p}\leq A, \forall \lambda>0\}\\&=\sup\{\lambda\big(\mu(|f|>\lambda)\big)^{1/p}: \lambda>0\}\tag{0}\label{zero}
\end{align}

Proof:
Denote by $a$ the left hand side of \eqref{zero} and by $\beta$ the righthand side.
As $f\in L^{\infty,p}$, there is $A>0$ such that
$$\lambda\big(\mu(|f|>\lambda)\big)^{1/p}\leq A, \qquad \lambda>0$$
Taking supremum over all $\lambda>0$ yields $\beta\leq A$. Taking the infimum over all such constants $A$ yields $\beta\leq \alpha$. On the other hand, for all $\lambda>0$, $\lambda\Big(\mu(|f|>\lambda)\Big)^{1/p}\leq\beta$. This means that $\alpha\leq \beta$.

Main result:
We state now the following important result:

Theorem 2: Suppose $f\in L^{\infty,p}(\mu)$, and let $E$ measurable with $\mu(E)<\infty$. For $0<q<p$, $f\mathbb{1}_{E}\in L_q(\mu)$ and
$$\int_E|f|^q\,d\mu\leq \frac{p}{p-q}\big(\mu(E)\big)^{1-\tfrac{q}{p}}[f]^q_p$$

Proof:
\begin{align}
\int_E|f|^q\,d\mu&=\int^\infty_0q \lambda^{q-1}\mu(E\cap\{|f|>\lambda\})\, d\lambda\\
&\leq \int^\infty_0q\lambda^{q-1}\min(\mu(E),\mu(|f|>\lambda))\,d\lambda\\
&\leq\int^\infty_0q\lambda^{q-1}\min(\mu(E),[f]^p_p\lambda^{-p})\,d\lambda\\
&=\mu(E)\int^{[f]_p/(\mu(E))^{1/p}}_0q\lambda^{q-1}\,d\lambda+[f]^p_p\int^\infty_{[f]_p/(\mu(E))^{1/p}}q\lambda^{q-1-p}\,d\lambda\\
&=\mu(E)\frac{[f]^q_p}{(\mu(E))^{q/p}} + \frac{q}{p-q}\frac{[f]^q_p}{(\mu(E))^{q/p-1}}=\frac{p}{p-q}(\mu(E))^{1-q/p}[f]^{q}_p
\end{align}
This implies that $f\mathbb{1}_E\in L_q(\mu)$ for all measurable $E$ of finite measure and that
$$
\newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 
    \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}\vertiii{f}_p:=\sup_{0<\mu(E)<\infty}\Big(\mu(E)\Big)^{\tfrac1p-\tfrac1q}\Big(\int_E|f|^q\,d\mu\Big)^{1/q}\leq\Big(\tfrac{p}{p-q}\Big)^{\tfrac1q}[f]_p
$$

Solution to the OP
We state the statement in the OP as a theorem.

Theorem 3:  Suppose $(X,\mathscr{B},\mu)$ is $\sigma$-finite. If $f\in L^{\infty,p}(\mu)$ and $0<q<p$, then
$$\newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 
    \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}[f]_p\leq\vertiii{f}_p$$

Proof: Suppose $X=\bigcup_nX_n$ with $0<\mu(X_n)<\infty$ and $X_n\subset X_{n+1}$ for all $n$. By Theorem 2,
$$\mu(X_n\cap\{|f|>\lambda\})\leq\frac{1}{\lambda^q}\int_{X_n\cap\{|f|>\lambda\}}|f|^q\le\frac{1}{\lambda^q}\mu(X_n\cap\{|f|>\lambda\})^{1-\tfrac{q}{p}}\newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 
    \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}
\vertiii{f}^q_p$$
Consequently
$$\big(\mu(X_n\cap\{|f|>\lambda)\big)^{\tfrac{q}{p}}\leq \newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 
    \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}\frac{\vertiii{f}^q_p}{\lambda^q}$$
Hence
$$\big(\mu(X_n\cap\{|f|>\lambda\})\big)^{1/p}\leq \newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 
    \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}\frac{\vertiii{f}_p}{\lambda}$$
The conclusion follows by letting $n\rightarrow\infty$.

Notes: $[\;]_p$ is not a norm, but it is almost like a norm

*

*For any scalar $c\neq0$,
\begin{align}
[cf]_p&=\sup_{\lambda>0}\{\lambda\Big(\mu(|c||f|>\lambda)\big)^{1/p}=\sup_{\lambda>0}\lambda\big(\mu\big(|f|>\tfrac{\lambda}{|c|}\big)^{1/p}\\
&=|c|\sup_{\lambda>0}\big|\tfrac{\lambda}{c}\big|\big(\mu\big(|f|>\tfrac{\lambda}{|c|}\big)^{1/p}=|c|[f]_p
\end{align}


*If $f,g\in L^{\infty,p}(\mu)$,
$$\mu(|f+g|>\lambda_1+\lambda_2)\leq\mu(|f|+|g|>\lambda_1+\lambda_2)\leq\mu(|f|>\lambda_1)+\mu(|f|>\lambda_2)$$
It follows that
$$\mu(|f+g|>\lambda)\leq \mu(|f|>\lambda/2)+\mu(|g|>\lambda/2)\leq 2^p\lambda^{-p}([f]^p_p+[g]^p_p)$$
Since $\sqrt[p]{|x|^p+|y|^p}\leq|x|+|y|$ for $p \ge 1$ and $\sqrt[p]{|x|^p+|y|^p}\leq2^{1/p-1}(|x|+|y|)$ for $p \le 1$,
$$[f+g]_p\leq 2([f]^p_p+[g]^p_p)^{1/p}\leq \max(2^{1/p},2)([f]_p+[g]_p)$$


*If $f\in L_p$, then $[f]_p\leq \|f\|_p$.
