Condition for $L^r(\mu)=L^\infty(\mu)$ where $\mu(X)=1$

I claim that if $\mu(X) = 1$ and $0 < r < s \leq \infty$, then the condition $L^r(\mu) = L^s(\mu)$ is equivalent to the existence of $c > 0$ such that $\mu(E) = 0$ or $\mu(E) \geq c$ for all measurable sets $E \subset X$.

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"$\Rightarrow$": Assume $L^r(\mu) = L^s(\mu)$. Then the linear map $$ \Phi : L^r(\mu) \to L^s(\mu), f \mapsto f $$ is well-defined. Furthermore, if $f_n \to f$ in $L^r$ and $f_n \to g$ in $L^s$, then also $f_n \to g$ in $L^r$, since (as you noted yourself) $L^s \hookrightarrow L^r$. This implies that $\Phi$ has closed graph. By the closed graph theorem (which also applies to quasi-Banach spaces; in fact, it applies to F-spaces), this means that $\Phi$ is a bounded linear map, say $\| f \|_{L^s} \leq C \cdot \| f \|_{L^r}$ for all $f \in L^r$. For $E \subset X$ measurable with $\mu(E) > 0$, this means $$ [\mu(E)]^{1/s} = \| 1_E \|_{L^s} \leq C \cdot \| 1_E \|_{L^r} = C \cdot [\mu(E)]^{1/r} , $$ and hence $[\mu(E)]^{\frac{1}{r} - \frac{1}{s}} \geq C$, so that finally $\mu(E) \geq C^{1/(\frac{1}{r} - \frac{1}{s})} =: c$, where we used that $\frac{1}{r} - \frac{1}{s} > 0$.

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"$\Leftarrow$": Assume that $\mu(E) = 0$ or $\mu(E) \geq c$ for all measurable $E \subset X$. It suffices to show (why?!) that $L^s (\mu) \hookrightarrow L^\infty(\mu)$. In fact, I claim that if $f \in L^s$, then $\| f \|_{L^\infty} \leq c^{-1/s} \| f \|_{L^s}$. Assume towards a contradiction that this is false; in particular, $\| f \|_{L^s} > 0$. Since $\| f \|_{L^\infty} > c^{-1/s} \| f \|_{L^s}$, there is $\lambda > c^{-1/s} \| f \|_{L^s}$ such that $E := \{ x : |f(x)| \geq \lambda \}$ satisfies $\mu(E) > 0$ and hence $\mu(E) \geq c$. This implies $$ \lambda \cdot c^{1/s} \leq \lambda \cdot [\mu(E)]^{1/s} = \| \lambda \cdot 1_{E} \|_{L^s} \leq \| f \|_{L^s} $$ and hence $\lambda \leq c^{-1/s} \| f \|_{L^s}$, contradicting our choice of $\lambda$.