Points In An Intervals Are Limit Points

I've given two solutions below. The first solution uses the supremum and infimum. The second solution doesn't use these words at all, but rather uses the same ideas needed to prove facts about the infimum and supremum of a set. Please read both proofs. Come with questions if you have any.

Solution using infimum and supremum of $(a,b)$

Pick $p\in M = [a,b] $. There are three cases to consider, namely $p=a$, $p=b$, and $p\in (a,b)$. We first suppose $p=a$. Let $I=(c,d)$ be an open interval that contains $p=a$. Since $a$ is the infimum of $(a,b)$, we know that any number larger than $a$ cannot be a lower bound of $(a,b)$. This means that $d$ is not a lower bound for $(a,b)$, so we can pick a number $x$ between $a$ and $d$ such that $x\in (a,b)$. Since $a<x<d$, we know $x\in I$ and $x\neq a$. Since $x\in (a,b)\subseteq [a,b] $, we know that $x\in M$. This completes that proof that $p=a$ is a limit point of $M$.

To prove that $p=b$ is a limit point of $M$, we use similar reasoning as above. Given an interval $I=(c,d)$ that contains $b$, we use the fact that $b$ is the supremum of $(a,b)$ to obtain a number $x\in (a,b)$ that lies between $c$ and $b$ (possible since $c$ is not an upper bound of $(a,b)$. We know $x\in M$ since $x\in (a,b)$. We also know $c<x<b$ which means $x\in I$ and $x\neq b$. This proves $p=b$ is a limit point of $M$.

To finish the proof, we now assume that $p\in (a,b)$ and must prove that $p$ is a limit point of $M$. Let $I=(c,d)$ be an open interval that contains $p$. We must produce a number $x$ such that $x\in I$, $x\in M$, and $x\neq p$. There are lots of ways to proceed, so what follows is not the only option. Let's look to the right of $p$. We know that both $b$ and $d$ are greater than $p$. All we need to do is pick a value for $x$ that is larger than $p$ but less than both $b$ and $d$. How do we do this? We use the fact that between any two real numbers, there is another real number. If $b<d$, then we pick $x\in (p,b)$. Otherwise, we know $d\leq p$ and we pick $x\in (p,d)$. So basically, we chose a number $x$ between $p$ and the smaller of $b$ and $d$. In either case, we have $p<x<b$ (hence $x\in M$) and $p<x<d$ (hence $x\in I$). Since $p<x$, we know $p\neq x$. This produces the needed value of $x$ to finish the proof that $p$ is a limit point of $M$.

Solution without infimum or supremum of $(a,b)$

Pick $p\in M = [a,b] $. There are two cases to consider, namely $p\in [a,b) $ and $p=b$. We first let $p\in [a,b)$. Let $I=(c,d)$ be an open interval that contains $p$. All we need to do is pick a value for $x$ that is larger than $p$ but less than both $b$ and $d$. How do we do this? We use the fact that between any two real numbers, there is another real number. Since both $d$ and $b$ are greater than $p$, we pick a number $x$ that is greater than $p$ and less than the smaller of $d$ and $b$. Since $c<p<x<d$, we know $x\in I$. Since $a\leq p<x<b$, we know $x\in M$. Since $p<x$, we know $p\neq x$. This completes that proof that $p\in [a,b)$ is a limit point of $M$.

To prove that $p=b$ is a limit point of $M$, we use similar reasoning as above, but this time pick a point left of $p=b$ rather than above it. Given an interval $I=(c,d)$ that contains $b$, we pick a number $x$ that is less than $b$ and greater than the larger of $a$ and $c$. Since $c<x<b$, we know $x\in I$. Since $a<x<b$, we know $x\in M$. Since $x<b$, we know $b\neq x$. This proves $p=b$ is a limit point of $M$.