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The question.

I was wondering whether or not the following assumption is true.

Let $E$ be a complete metric set containing a point verifying the property $\mathcal P$. Then $E$ is uncountable.

A point $x\in E$ verifies the property $\mathcal P(x)$ if for every neighbourhood $V$ of $x$, there exists $y\in V\setminus \{x\}$ and $y$ verifies $\mathcal P(y)$.

In other words, $\mathcal P(x)$ is true if, and only if, $x$ is not isolated, and for every neighbourhood $V$ of $x$, $V\setminus \{x\}$ contain a non isolated point verifying the same property.

What I did.

I tried to prove the veracity of the result since I wasn't able to find a counterexample.

Let $x\in E$ be a point such that $\mathcal P(x)$ is true.

Then we can construct recursively a sequence $(x_n)$:

$$\begin{cases} x_0\ne x \\ \mathcal P(x_0)\text{ is true.}\end{cases}$$

Then assuming $x_0,\ldots,x_n$ are constructed, let's define $r_n$ such that

$$\forall i\in\{1,\ldots,n-1\},\quad x_i\notin B(x_n,r_n).$$

We have a set $X=\{x_i\}$ of points of $E$. Since when can always double the size of $X$ because every $x_i$ verifies $\mathcal P(x_i)$, I would tend to think that we can construct a set $Y$ in bijection with $2^X$, which would have the cardinality $2^{\aleph_0}$, so $Y$ would be uncountable. So $E$ would be.

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    What's a complete set? Also isn't your condition just stating that no point is isolated? – 2017-01-08
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    @AlessandroCodenotti Not exactly. I edited to specify what I meant. – 2017-01-08
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    I see, it's still unclear to me what complete means in a generic (non-metrizable) context – 2017-01-08
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    @AlessandroCodenotti You are right, I will edit. I assume $E$ metric. – 2017-01-08
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    Your implicit definition of property $\mathcal P$ is confusing. Nevertheless, if we define $P=\{x\in E:\mathcal P(x)\},$ then it seems clear that $P$ is a nonempty closed set with no isolated points, i.e., it's a perfect set. If $E$ is a complete metric space, then $P$ must have at least the cardinality of the continuum. To see this directly, start by choosing a point $x\in P$ and a neighbourhood $U$ of $x$ of diameter $\lt1.$ Next find two neighbourhoods $U_0,U_1$ inside $U,$ each containing a point of $P,$ and each of diameter $\lt\frac12.$ And so on. – 2017-01-08

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Let $A = \{ x \in E : \mathcal{P}(x)\}$. By assumption, $A \neq \varnothing$. By definition of $\mathcal{P}$, every $x\in A$ is a limit point of $A$. So $A$ is infinite. Let $P = \overline{A}$ (note: since $A = E' \cap E''$, and derived sets are closed in $T_1$-spaces, in fact $P = A$). Then $P$ is a perfect subset of $E$, hence uncountable.