If $0 \in P$ and $\sup\{x \in [0,1] : [0,x] \subseteq P\} = 1$, then $[0,1] \subseteq P$? (What if adding 3rd assumption?)

Question

Let $P \subseteq \mathbb{R}$. If $0 \in P$ and $\sup\{x \in [0,1] : [0,x] \subseteq P\} = 1$, then $[0,1] \subseteq P$.

Is the above claim true or false? How to prove or disprove it?

Edit1: Thanks for the quick disproof/counterexample! In fact this question originated from reading a proof that a closed and bounded interval in $\mathbb{R}$ is compact. If adding 3rd assumption:

$\sup\{x \in [0,1] : [0,x] \subseteq P\} \in \{x \in [0,1] : [0,x] \subseteq P\}$

can this stricter claim be proven to be true?

Answer 1

(Answering the second, post-edit question).

Let $P \subseteq \mathbb{R}$. Assume

1. $0 \in P$
2. $\sup\{x \in [0,1] : [0,x] \subseteq P\} = 1$.
3. $\sup\{x \in [0,1] : [0,x] \subseteq P\} \in \{x \in [0,1] : [0,x] \subseteq P\}$.

Assumption 2 substitutes into Assumption 3 to say that $1 \in \{x \in [0,1] : [0,x] \subseteq P\}$ which implies $1 \in [0,1]$ and $[0,1] \subseteq P.$

Answer 2

This is false : for $P=[0,1)$, you have that $\sup\{x\in[0,1];[0,x]\subset P\}=\sup[0,1)=1.$ Thus, $1\in[0,1]$ and $1\notin P$.