$\newcommand{\ee}{ \varepsilon }$ Let $(X,d)$ be a metric space. A subset $P\subset X$ is perfect if $P$ is the set of its accumulation points. I want to prove that if $P$ is perfect then $|P|\ge \mathfrak{c}$. Moreover, I want to prove that if $(X,d)$ is not complete then the above result is not generally true.
Here's how I'm trying to prove this (where I'm marking in yellow background those parts which about which I'm in a doubt):
Let $P\subset X$ be a perfect set. Let $x,y\in P$ be two points satisfying
$$\mbox{diam}(P)=\sup\limits_{x,y\in P}\mbox{d}(x,y)$$
Then $\mbox{diam}(P)\ne 0$, since all points in $P$ are accumulation points, so any closed ball $B[x,\varepsilon]$ contains infinitely many points of $P$.
[In a doubt because may be lacking rigour].
Let $0< \ee<\frac{\mbox{diam}(P)}{4}$ and define $r_1 := \frac{\mbox{diam}(P)}{4}-\ee$. Then let $x_0 \in P$ be such that $d(x_0,x)=\frac13d(x,y)$ and $d(x_0,y)=\frac23d(x,y)$. Let $x_1 \in P$ be such that $d(x_1,y)=\frac13d(x,y)$ and $d(x_0,x)=\frac23d(x,y)$. Then the following holds for these closed balls, by construction:
$$B[x_0,r_1]\cap B[x_1, r_1]=\emptyset$$
[In a doubt because I don't see how to formally define a ball that is guaranteed to be contained in $P$].
Now let $0<\ee'<\frac{r_1}{2}$ and define $r_2:=\frac{r_1}{2}-\ee'$. Let $b_i\in \{0,1\}$. Let $x_{b_10}$ be two points in the set
$$S_{b_10} = \{x\in X: d(x,x_0)=r_2\}$$ such that $d(x_{00}, x_{10})=2r_2$.
Also let $x_{b_11}$ be two points in the set
$$S_{b_11} = \{x\in X: d(x,x_1)=r_2\}$$ such that $d(x_{01}, x_{11})=2r_2$.
Then $B[x_{b_1b_2},r_2]\subset B(x_{b_1},r_1)\setminus\{x_{b_1}\}$ and $B[x_{b_10}, r_2]\cap B[x_{b_11},r_1]=\emptyset$.
Inductively, let $0<\ee^{(n)}<\frac{r_n}{4}=\frac{r_1}{2^{n+1}}$ and let $x_{b_1\dots b_n 0}$ be the points in the set
$$S_{b_1\dots b_n 0}=\{x\in X: d(x, x_{b_1\dots b_{n-1} 0})=r_{n+1}\}$$
satisfying $d(x_{b_1\dots 00}, x_{b_1\dots 10})=2r_{n+1}$.
And let $x_{b_1\dots b_n 1}$ be the points in the set
$$S_{b_1\dots b_n 1}=\{x\in X: d(x, x_{b_1\dots b_{n-1} 1})=r_{n+1}\}$$
satisfying $d(x_{b_1\dots 01}, x_{b_1\dots 11})=2r_{n+1}$.
Let $r_{n+1}=r_n-\ee^{(n)}$
Then $B[x_{b_1\dots b_n b_{n+1}},r_{n+1}]\subset B(x_{b_1\dots b_n },r_n)\setminus\{x_{b_1\dots b_n }\}$ and $B[x_{b_1\dots b_n 0}, r_{n+1}]\cap B[x_{b_1\dots b_n 1},r_{n+1}]=\emptyset$.
We thus obtain the following set of sequences:
$$\chi=\{(x_{b_1}, x_{b_1b_2}, x_{b_1b_2b_3},\dots)\subset P: b_n\in \{0,1\}, n\in\mathbb N\}$$
By the Cantor diagonal argument, $\chi$ is an uncountable set of sequences. Let $x_d\in \chi$ and consider now the following map:
$$F(x_d):\chi \to P$$ defined as
$$F(x_d) =\lim\limits_{n\to\infty} x_d= \lim\limits_{n\to\infty} (x_{b_1}, \dots, x_{b_1\dots b_n})$$
Since $(X,d)$ is complete and $P\subset X$, with each $x_d$ being a sequence of closed balls contained in $P$, $x_d$ converges in $P$.
Moreover, $F$ is an injection.
[How does one formally prove this?]
So we have that $|P|\ge |\chi|\ge |\mathbb R|=\mathfrak{c}$.
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Finally, let $X:=(a,b)\cap \mathbb Q$ be an interval in $\mathbb{Q}$. Then $X$ is perfect. But $|X|\le |\mathbb{Q}|=\aleph_0<\mathfrak{c}$.
Would appreciate if you could please help me with those parts marked with yellow. Also, please let me know if my proof is fine other than those parts.
