Confusion about the uniqueness (up to isomorphism) of $(\Bbb R,<)$

Question

This theorem is from textbook Introduction to Set Theory by Karel Hrbacek and Thomas Jech

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5.7 Theorem $(\Bbb R,<)$ is the unique (up to isomorphism) complete linearly ordered set without endpoints that has a countable subset dense in it.

Proof. Let $(C, \prec)$ be a complete linearly ordered set without endpoints, and let $P$ be a countable subset of $C$ dense in $C$. Then $(P, \prec)$ is isomorphic to $(\Bbb Q, <)$ and by the uniqueness of completion (Theorem 5.3), $(C, \prec)$ is isomorphic to the completion of $(\Bbb Q, <)$, i.e., to $(\Bbb R,<)$.

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First, let me present relevant theorems that lead to the above theorem.

1. Let $(P,<)$ and $(Q,\prec)$ be countable, dense, and linearly ordered sets without endpoints. Then $(P,<)$ and $(Q,\prec)$ are order-isomorphic.

2. Let $(P, <)$ be a dense linearly ordered set without endpoints. Then there exists a complete linearly ordered set $(C,\prec)$ such that

(a) $P\subseteq C$.

(b) If $p,q\in P$ then $p < q$ if and only if $p \prec q$ ($<$ coincides with $\prec$ on $P$).

(c) $P$ is dense in $C$, i.e., for any $a,b\in C$ such that $a\prec b$, there is $p\in P$ with $a\prec p \prec b$.

(d) $C$ does not have endpoints.

Moreover, this complete linearly ordered set $(C,\prec)$ is unique up to isomorphism over $P$. In other words, if $(C^\ast,\prec^\ast)$ is a complete linearly ordered set which satisfies (a)-(d), then there is an isomorphism $h$ between $(C,\prec)$ and $(C^\ast,\prec^\ast)$ such that $h(x) = x$ for each $x \in P$. The linearly ordered set $(C,\prec)$ is called the completion of $(P, <)$.

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My questions:

1. The proof of theorem 5.7 said that let $P$ be a countable subset of $C$ dense in $C$. Then $(P, \prec)$ is isomorphic to $(\Bbb Q, <)$

But from (1), we can see that in order for $(P, \prec)$ and $(\Bbb Q, <)$ to be isomorphic, $P, Q$ must be countable, dense, linearly ordered, and without endpoints. I'm unable to understand where in the proof P is mentioned to be dense ($P$ must be dense itself, not just dense in $C$), and without endpoints.

2. I think the authors use the following argument: $(\Bbb R,<)$ is the completion of $(\Bbb Q,<)$, $(C,\prec)$ is the completion of $(P,\prec)$, and $(\Bbb Q,<)$ and $(P,\prec)$ are isomorphic from (1). Then $(\Bbb R,<)$ and $(C,\prec)$ are isomorphic from (2).

I'm unable to understand why $(\Bbb R,<)$ and $(C,\prec)$ are isomorphic, please explain!

Thank you so much!