First we prove this lemma: given any $\newcommand{\eps}{\epsilon} \newcommand{\ma}{\mathbb} \newcommand{\nf}{\infty} \newcommand{\fa}{\;\forall\;}\eps>0$, there exists some finite collection of points $x_1,x_2,\dots,x_n\in X$ such that the finite collection of open balls $\{B(x_i;\eps)\}_{i=1}^n$ covers $X$. To prove this, we show the contrapositive. We suppose this is not the case, and then show that $X$ has an infinite set with no limit point (this property is called limit point compactness, by the way).
Letting $y_1$ be any point of $X$, $B(y_1;\eps)$ cannot cover $X$ (because otherwise $\{B(y_1;\eps)\}$ would be our desired finite open covering). Now suppose we are given points $y_1,y_2,\dots,y_m$ such that the collection $\{B(y_i;\eps)\}_{i=1}^m$ does not cover $X$. We then set $y_{m+1}$ equal to any point of the nonempty set $X-\cup_{i=1}^m B(y_i;\eps)$. In this way, we have constructed an infinite sequence of points, $\{y_i\}_{i=1}^{\nf}$, such that for any integer $n\in\ma Z^+$, the collection $\{B(y_i;\eps)\}_{i=1}^n$ does not cover $X$. Because of the way we constructed this sequence of $y_i$s, it is clear that for any $j>k$, $y_j$ does not belong to $B(y_k;\eps)$, so $d(y_j,y_k)\geq \eps$. Interchanging the roles of $j$ and $k$, we get that $d(y_j,y_k)\geq \eps, \fa j\neq k$. We now have everything we need to demonstrate a contradiction. I claim that $A=\{y_i\}_{i=1}^{\nf}$ is an infinite set with no limit point. because $d(y_j,y_k)\geq\eps, \fa j\neq k$, $y_j\neq y_k, \fa j\neq k$, so all $y_i$s are distinct, making $A$ infinite. To show that $A$ has no limit point, suppose to the contrary that it does. Call this limit point $z$. As $z$ is a limit point of $A$ and $X$ (being a metric space) is $T_1$, the neighborhood $B(z;\frac{\eps}{2})$ contains infinitely many points of $A$. This means, in particular, that we choose distinct integers, $p$ and $q$, such that $y_p$ and $y_q$ both belong to $B(z;\frac{\eps}{2})$. But this means $d(y_p,y_q)\leq d(y_p,z)+d(z,y_q)<\frac{\eps}{2}+\frac{\eps}{2}=\eps$, a contradiction. It follows that $A$ has no limit point.
Taking the contrapositive of what was just proved, we get the statement of the first paragraph (assuming $X$ is a limit point compact metric space). Now for each positive integer, $n$, let $\{x_{i,n}\}_{i=1}^{m_n}$ be a finite collection of points in $X$ such that $\{B(x_{i,n};\frac{1}{n})\}_{i=1}^{m_n}$ covers $X$. The collection of points $\{x_{i,n}\}_{1\leq i\leq m_n, 1\leq n\leq \nf}$ is clearly countable. I now leave it to you to show that this collection is also dense in $X$.
(a fact that is relevant and interesting here is that the kinds of metric spaces your question is about are exactly the compact metric spaces)
Best Answer
Pick some $x \in A$, and consider $A\setminus \{x\}$. This is a proper subset of $A$, and therefore countable. Now consider $A\setminus \{x\} \cup \{x\}$, which is countable as it's the union with a singleton. But this is the same as $A$, so $A$ is countable.