Exercise: Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers. Let $L^+ = \lim \sup_{n\to\infty} (a_n)_{n=m}^\infty$ and $L^-=\lim \inf_{n\to\infty}(a_n)_{n=m}^\infty$. Prove that $$\inf(a_n)_{n=m}^\infty \le L^-\le L^+\le \sup(a_n)_{n=m}^\infty$$ I have already proven that $\inf(a_n)_{n=m}^\infty\le L^-$ and $L^+\le \sup(a_n)_{n=m}^\infty$. This just follows from the definitions and the following proposition. I am not sure if my proof regarding $L^-\le L^+$ is correct or not. I would appreciate if someone could check whether it is correct or not.
Proposition $1$: Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers. If $y<\sup(a_n)_{n=m}^\infty$, then there exists an $n$ such that $y<a_n\le \sup(a_n)_{n=m}^\infty$. Similarly if $y>\inf(a_n)_{n=m}^\infty$, then there exists an $n$ such that $y>a_n\ge \inf(a_n)_{n=m}^\infty$.
Notation: The sequence $(a^+ _N)_{n=m}^\infty$ is the sequence defined by the formula $$a^+ _N :=\sup(a_n)_{n=N}^\infty$$
The sequence $(a^- _N)_{N=m}^\infty$ is the sequence defined by the formula $$a^- _N:=\inf(a_n)_{n=N}^\infty$$ These are essentially the sequences from where the definition of limit superior and limit inferior come from. I think different books use different notation. This is the one used in the book I am reading. So, the book defines $$\lim \sup_{n\to\infty}(a_n)_{n=m}^\infty := \inf(a^+ _N)_{N=m}^\infty$$ $$\lim\inf_{n\to\infty} (a_n)_{n=m}^\infty := \sup(a^- _N)_{N=m}^\infty$$
The following is the proof of $L^-\le L^+$.
Proof: First we prove that $a^+ _N\ge a^- _N$ for all $N$. By definition we have that $a^+_N=\sup(a_n)_{n=N}^\infty$ and $a^- _N=\inf(a_n)_{n=N}^\infty$. Thus, we need to prove that $\sup(a_n)_{n=N}^\infty \ge \inf(a_n)_{n=N}^\infty$. If $\sup(a_n)_{n=N}^\infty=\inf(a_n)_{n=N}^\infty$, then we are done. So suppose for the sake of contradiction that $\inf(a_n)_{n=N}^\infty >\sup(a_n)_{n=N}^\infty$. Then by definition of supremum we have that for all $n$, $$\inf(a_n)_{n=N}>\sup(a_n)_{n=m}^\infty \ge a_n$$ Which contradicts the definition of the infimum of a sequence. Thus, $$\sup(a_n)_{n=N}^\infty \ge \inf(a_n)_{n=N}^\infty$$
Now to prove that $L^-\le L^+$, we have to show that $$\sup(a^-_N)_{N=m}^\infty \le \inf(a^+_N)_{N=m}^\infty $$ If they are both equal, then we are done. So, suppose for the sake of contradiction that $$\sup(a^-_N)_{N=m}^\infty > \inf(a^+_N)_{N=m}^\infty$$ Then by proposition $1$, there exists an $N$ such that $$\sup(a^-_N)> a^+_N\ge \inf(a^+_N)_{N=m}^\infty$$ Then again by proposition $1$ there exists an $N'$ such that $$\sup(a^-_N)_{N=m}^\infty \ge a^-_{N'}>a^+_N\ge \inf(a^+_N)_{N=m}^\infty$$ But this contradicts the previous result that $a^+_N\ge a^-_N$ for all $N$. Thus, we have that $$L^-\le L^+$$
Is this proof correct? If so, is there a simpler way to prove this inequality?
Your proof is correct. Whether or not you can simplify it depends on if and how you've defined suprema/infima for subsets of ordered fields.