I am reading "A Course in Analysis vol.3" by Kazuo Matsuzaka.
The author proved the following proposition as follows:
An interval in $\mathbb{R}$ is connected
Proof:
Let $I$ be an interval in $\mathbb{R}$.
We assume that $I = P \cup Q$, $P \cap Q = \emptyset$, $P, Q$ are both non-empty and open in $I$.
Let $a \in P$ and $b \in Q$.
Since $a \ne b$, $a < b$ or $a > b$.
Without loss of generality, we can assume that $a < b$.
Since $I$ is an interval, $[a, b] \subset I$.
Let $M := [a, b] \cap P$.
Since $a \in M$, $M \ne \emptyset$. $b$ is an upper bound of $M$.
So, there exists $c := \sup M$ and $a \leq c \leq b$.
Since $c \in I$, $c \in P$ or $c \in Q$.
If $c \in P$, $c < b$.
Since $P$ is open in $I$, $P = P' \cap I$ for some set $P'$ which is open in $\mathbb{R}$.
So, $c < c + \epsilon < b, c + \epsilon \in P$ for sufficiently small $\epsilon > 0$.
$\cdots$
I think we don't need the sentence "Since $P$ is open in $I$, $P = P' \cap I$ for some set $P'$ which is open in $\mathbb{R}$." in the proof.
Is the above sentence really necessary for the proof?
My proof without the above sentence is here:
Proof:
Let $I$ be an interval in $\mathbb{R}$.
We assume that $I = P \cup Q$, $P \cap Q = \emptyset$, $P, Q$ are both non-empty and open in $I$.
Let $a \in P$ and $b \in Q$.
Since $a \ne b$, $a < b$ or $a > b$.
Without loss of generality, we can assume that $a < b$.
Since $I$ is an interval, $[a, b] \subset I$.
Let $M := [a, b] \cap P$.
Since $a \in M$, $M \ne \emptyset$. $b$ is an upper bound of $M$.
So, there exists $c := \sup M$ and $a \leq c \leq b$.
Since $c \in I$, $c \in P$ or $c \in Q$.
If $c \in P$, $c < b$.
If $c < x < b$, then $x \in I$.
$P$ is open in $I$.
So, there exists $\epsilon' > 0$ such that if $c < x < c + \epsilon' < b$ and $x \in I$, then $x \in P$.
But if $c < x < b$ then $x \in I$.
So, there exists $\epsilon' > 0$ such that if $c < x < c + \epsilon' < b$, then $x \in P$.
Let $c < c + \epsilon < c + \epsilon'$.
Then, $c < c + \epsilon < b$ and $c + \epsilon \in P$.
$\cdots$
It feels like that sentence is trying to justify the next one you write: So, $c<c+\varepsilon<b$, $c+\varepsilon \in P$, for sufficiently small $\varepsilon$.
As long as this is clear from you from the fact that $P$ is open in $I$ then it's all good and you don't need that sentence... unless there is something else later in the proof that appears to be using it too.