Proving Subspace of a Separable Metric Space is Separable: Seeking Direct Proof

Question

Remark:

I am aware that in a metric space, the existence of a countable base is equivalent to separability. Therefore, one could argue(Every subspace of a separable metric space is separable) that since $X$ has a countable base, $Y$ must also have a countable base (as the open sets in $Y$ are just intersections of open sets in $X$ with $Y$), and hence $Y$ is separable. However, this proof seems indirect, and I am interested in finding a more direct proof.

Question:

Let $(X, d)$ be a separable metric space with a countable dense subset $Z$. Prove that any subspace $Y$ of $X$ is also separable.

My Attempt:

Difficulty:

For some points $p$ in $Y$, it's not straightforward to find a sequence in $Y \cap Z$ that converges to $p$.

Solution:

We consider two cases:

Case 1: $p$ is an isolated point in $Y$

In this case, we can find an open set $U_p$ in $X$ such that $U_p \cap Y = \{ p \}$. Since $Z$ is dense in $X$, there exists a point $q \in Z$ such that $q \in U_p$. This defines an injection $i: S \to Z$, where $S$ is the set of isolated points in $Y$. Therefore, $S$ is at most countable.(afterwards i plan to combine $S$ with $Z\cap Y$, which is also at most countable)

Case 2: There exists a sequence $\beta$ in $Y$ converging to $p$

We claim that there exists another sequence $\alpha$ in $Y \cap Z$ that also converges to $p$.

If this is not the case, we can find a ball $B(p, r_1)$ such that $B(p, r_1) \cap Y \cap Z = \emptyset$. Let $z_1, \ldots, z_k$ be the points of $\alpha$ inside $Y$ and let $d_1, \ldots, d_k$ be their distances to $p$. Without loss of generality, assume $d_1$ is the smallest and $d_1 \geq r_1$. Set $d = \frac{r_1}{3}$.

Case 2.1: $\beta$ contains an interior point $y$ of $Y$ with $y \neq p$ and $y \in B(p, d)$

In this case, we can find an open set $U$ contained in $Y$ and $B(p, d)$. However, $U \cap Z = \emptyset$, which is a contradiction.

Case 2.2: All points in $\beta$ are boundary points of $Y$

(Here, I got stuck.)

Answer 1

Since $Y\cap Z$ may be empty, you cannot directly argue that it is dense in $Y$. This will not be the case in many examples, e.g, take $X=\mathbb R$, $Z=\mathbb Q$, and $Y=\mathbb R\backslash \mathbb Q$.

So for case two your claim is not correct.

Moreover, in case 1 you would need to do a little more work to guarantee $i$ is an injection, as your current construction allows $U_{p_1}$ and $U_{p_2}$ to overlap.

In general, if you want a somewhat direct proof (or any proof), you will almost certainly need a countable version of the axiom of choice per this result.

Proof

Once we bite the bullet and agree to use countable choice, we can choose a point $y_{z,k}$ from each set $B(z,\frac{1}{k})\cap Y$ whenever $B(z,\frac{1}{k})\cap Y$ is nonempty. Then let $$S=\{ y_{z,k}\mid z\in Z, k\in\mathbb N, B(z,\frac{1}{k})\cap Y\neq \emptyset\}\text{.}$$ This gives a countable union of countable sets which is therefore countable.

Now sometimes $y_{z,k}$ is undefined, however, for each $y\in Y$ we may take a sequence $z_n\to y$ by density of $Z$, assuming wlog that $d(z_n,y)<\frac{1}{k_n}$ for some $k_n\to \infty$, and thus in this case, $y_{z_n,k_n}$ is indeed defined, and we have $$d(y_{z_n,k_n},y)\leq d(y_{z_n,k_n},z_n)+d(z_n,y)\leq \frac{2}{k_n}\to 0\text{,}$$ so $y\in\bar{S}$.