Definition. Let $(E,d)$ be a metric space.
- A base of open sets for $E$ is a family $\mathscr{B}$ of open subsets of $E$ such that, for any open subset
$O$ of $E$, there is collection $\{O_i\}_{i\in I}$ contained in
$\mathscr{B}$ such that $$O=\bigcup_{i\in I} O_i.$$
- If $F\subseteq E$, when we say that $F$ is a subspace of $E$, we are considering the metric space $(F,\rho)$, where
$$\rho=d\big|_{F\times F}.$$
It is a good exercise to prove (one direction is provided in a Brian's comment) that:
Theorem 1. Let $E$ be a metric space. $E$ is separable if and only if $E$ has a countable base of open sets.
From this follows:
Corollary 2. Let $E$ be a metric space. $E$ is separable if and only if all its subspaces are separable.
Proof. Assume first that $E$ is separable. By Theorem 1, there exist a countable base of open sets $\mathscr{B}$. Let $F$ a subspace of $E$. Then the family $\mathscr{B}'=\{O\cap F : O\in \mathscr{B}\}$ is countable base of open sets for $F$. By Theorem 1 we conclude $F$ is separable.
Conversely, suppose that $E$ is such that all its subspaces are separable. $E$ is a subspace of $E$, so $E$ is separable.
From Theorem 1 we also get:
Lema 3. Let $E$ be a separable metric space. If $\{U_i\}_{i\in I}$ is a family of pairwise disjoint, open, nonempty sets, then $I$ is countable.
Proof. Since $E$ is separable we can assume that there is a base of open sets $\mathscr{B}=\{O_j\}_{j\in\Delta}$ with $\Delta\subset \Bbb N$.
Since the open sets in $\{U_i\}_{i\in I}$ are pairwise disjoint and nonempty, for each $i\in I$, you can pick (axiom of choice possibly needed here) an element $x_i\in U_i$. Notice that the $\{x_i\}_{i\in I}$ are necessarily pairwise distinct.
Because $\mathscr{B}$ is a base of open sets, for each $i\in I$ there exist a $j_i\in\Delta$ such that
$$x_i\in O_{j_i}\subseteq U_i.$$
Notice that $i\neq k$ implies $j_i\neq j_k$. Otherwise there would be a pair of open sets in $\{U_i\}_{i\in I}$ whose intersection is not empty.
Then, the function $f:I\to\Delta$ given by
$$f(i)=j_i$$
is an injection from $I$ to $\Delta$, a countable set. Therefore $I$ must be countable.
Now, the answer to your question:
Theorem 4. Let $E$ be a metric space. If $E$ is separable and $A$ is a subset of $E$ such that all its points are isolated, then $A$ is countable.
Proof. Consider $A$ as subspace of $E$. By Corollary 2, $A$ is separable.
Since all the points of $A$ are isolated, for each $x\in A$ there exist a $r_x\gt 0$ such that
$$B(x,r_x)\cap A=\{x\}.$$
Thus
$$\{B(x,r_x)\cap A\}_{x\in A}$$ is a family of pairwise disjoint open subsets of $A$. Since $A$ is separable, by Lemma 3, this family must be countable. Since there is exactly one set in that family for each element of $A$,$A$ must be countable.
Finally, it is worth to point out that
Corollary 5. Let $E$ be a metric space. If there exist a uncountable subset $A$ of $E$ such that all its points are isolated then $E$ is not separable
Just so there is the option to remove this question from the unanswered queue, here is a post consolidating the info in the comments above.
A key thing to grasp is that, for general topological spaces, separability is not a hereditary property (a topological property is called hereditary if, whenever a topological space $X$ has the property, so does every subspace $Y \subseteq X$). Here is an example showing that separability is not necessarily inherited by a subspace.
Example: Let $I$ denote the closed unit interval and let $X = I^I$, the product of continuum-many copies of $I$. The space $X$ is separable and, by Tychonoff's theorem, a compact Hausdorff space to boot. An example of a countable dense subset $D \subseteq X$ is the set of all rational valued functions $I \to I$ which are piecewise-constant with respect to a rational partition of $I$. Another example of a countable dense subspace is the set of all maps $I \to I$ which are described by a polynomial with rational coefficients. An example of a nonseparable subspace $Y \subseteq X$ is the set of all characteristic functions of points. That is, the set of all functions $\chi_p : I \to I$, $p \in I$, given by $\chi_p(x) = 0$ if $x \neq p$, and $\chi_p(p) = 1$. The subspace topology on $Y$ is discrete, so the only dense subspace of $Y$ is $Y$ itself. Since $Y$ is also uncountable, it is not separable.
Having digested the example above, or some alternative, one understands that the statement to be proved must utilize, in a key way, the fact that the $X$ in the problem is a metric space. One way to proceed is to establish, and combine, following standard facts:
Fact 1: A separable metric space is second-countable.
Fact 2: A subspace of a second-countable topological space is second countable. In other words, second-countability is a hereditary property.
Fact 3: A second-countable topological space is separable.
Proofs of the above facts can be found in any introduction to point-set topology. However, the best course of action would be to attempt to prove them yourself first.
Best Answer
Let $C$ be the set of isolated points of $Y$. We will define an injective map, from $C$ to $A$, the countable dense set in $X$. The existence of such a map implies that $C$ is countable.
In order to do so, recall that for all $x\in C$ there is an $r_x>0$ s.t. $B_{r_x}(x)\cap Y=\{x\}$. Now let $f(x)$ be any point in $B_{r_x/2}(x)\cap A$, which is non empty by the density of $A$.
Now we are left to show the injectivity of $f$, let $x,y\in C$ s.t. $f(x)=f(y)$. Without loss of generality $r_x\geq r_y$. Then $f(y)\in B_{r_x/2}(x)$ and at the same time $f(y)\in B_{r_y/2}(y)$, hence $y\in B_{r_x}(x)\cap Y=\{x\}$. This is $x=y$. Hence $f$ is injective.