$\textbf{Theorem}$ (Bertrand-Chebyshev theorem/Bertrand's postulate): For all integers $n\geq 2$, there exists an odd prime number $p\geq 3$ satisfying $n<p<2n$.
$\textit{Proof }$: For $n=2$, we can simply take $p=3$ which is of course an odd prime number, so assume henceforth that $n\geq 3$. Now make the strong inductive hypothesis that the Bertrand-Chebyshev theorem is true for all integers $2\leq m<n$. In particular, since $n\geq 3\Rightarrow 2\leq n-1<n$, we see that there exists an odd prime number $p\geq 3$ satisfying $n-1<p<2(n-1)\Rightarrow n\leq p<2n$. We now divide into $2$ cases:
$\textbf{Case }1$: If $n\neq p$, then we can improve the above inequality to say $n<p<2n$, which would mean that we're done.
$\textbf{Case }2$: If $n=p$, then assume for sake of contradiction that all of the integers $p'$ satisfying $p<p'<2p$ are composite numbers. Now construct a well-defined function $f:\{p+1,p+2,...,2p-2,2p-1\}\to\{2,3,...,p-2,p-1\}$ which maps each of those $p-1$ composite numbers to their respective $\textit{smallest prime factor}$. To elaborate a little more on why $f$ is well-defined, the reason that $f(p')<p$ for all integers $p'$ satisfying $p<p'<2p$ is because first, $f(p')\leq p'$ because it is a $\textit{factor}$ of $p'$, second $f(p')\notin\{p+1,p+2,...,p'\}$ because it is a $\textit{prime}$ factor of $p'$ (while all those integers were by assumption composite numbers), and finally $f(p')\not = p$ because $p<p'<2p\Rightarrow 1<p'/p<2$, so $p'/p$ cannot possibly be an integer, hence $p$ cannot possibly be a factor of $p'$, let alone be the smallest prime factor $f(p')$ of $p'$.
Continuing in Case $2$ above, we remark that the codomain $\{2,3,...,p-2,p-1\}$ of $f$ has only $p-2$ integers in it while $f$'s domain of composite numbers $\{p+1,p+2,...,2p-2,2p-1\}$ has $p-1>p-2$ elements in it. Therefore, by the pigeonhole principle, there must be two composite numbers $a,b$ satisfying $p<a<b<2p$ such that $f(a)=f(b)$. Let us denote this prime number as $q:=f(a)=f(b)$. From the discussion in the preceding paragraph, we see that $2\leq q<p$, so by the original strong inductive hypothesis that we made, we see that there exists an odd prime number $r\geq 3$ satisfying $q<r<2q$. Now within this Case $2$, we have to break into $3$ more subcases and show that in each of the $3$ subcases, a contradiction arises.
$\textbf{Subcase }1$: If $r>p$, then since $r<2q<2p$, we see that we would have $p<r<2p$, which would directly give a contradiction to the original assumption that all the integers $p'$ satisfying $p<p'<2p$ are composite.
$\textbf{Subcase }2$: If $r=p$, then we would have $q<p<2q\Rightarrow p<2q<2p\Rightarrow 1/p<1/q<2/p$. Recall that we also had $p<a<2p$ and $p<b<2p$, so multiplying each of these inequalities with the one above, we have: $$1<a/q<4$$ $$1<b/q<4$$ But since by construction $q$ was a factor of $a$ and was also a factor of $b$, we see that $a/q$ and $b/q$ must both be integers, so in fact we can improve the above inequalities to say: $$2\leq a/q\leq 3$$ $$2\leq b/q\leq 3$$ This means that $a\in\{2q,3q\}$ and similarly $b\in\{2q,3q\}$. But recalling that $a<b$ and $q\geq 2>0\Rightarrow 3q>2q$, we in fact deduce that $a=2q$ and $b=3q$. Since $q$ was also the $\textit{smallest prime}$ factor of $a$, and $2$ is clearly seen to be a prime factor of $a$, we conclude that we must have $q=2$, hence $a=4$ and $b=6$. But recalling the inequalities $p<a$ and $b<2p$, we see this would imply $p<4$ and $p>3$, but there are no integers strictly between $3$ and $4$.
$\textbf{Subcase }3$: If $r<p$, then we would just apply the strong inductive hypothesis yet again to locate an odd prime number $s\geq 3$ satisfying $r<s<2r$. If $s\geq p$, then we could reason as in the above $2$ subcases, and if $s<p$, then we would just apply the strong inductive hypothesis $\textit{yet again}$, doing it as many times as necessary until we obtain an odd prime number which is greater than or equal to $p$, and hence handled by the above $2$ subcases (note that this algorithm would be guaranteed to terminate because otherwise we would have a strictly monotone increasing sequence of odd prime numbers that is bounded from above by $p$).