Let $\varepsilon$ denote the Euclidean topology on $\mathbb R$.

**Proposition:** *There is a 0-dimensional $T_2$ topology $\tau$ on $\mathbb R$
such that $\tau$ and $\varepsilon$ intersect in only the co-finite sets.*

*Proof:*
We need the following lemma:

**Lemma:**
*Let $Y=(Y,\nu)$ be an infinite topological space such that
$$|\overline{E}^\nu|\ge 2^{\omega}$$ for all $E\in [Y]^\omega$.
Then there is an injective map $f:\mathbb R\to Y$
such that
$$
\forall D\in [\mathbb R]^\omega \text{ if }\overline D\ne \mathbb R
\text{ then }
\exists x_D\in (\mathbb R\setminus \overline D) \ f(x_D)\in \overline{f[D]}^\nu.
$$*

*Proof of the lemma:*

Enumerate $[\mathbb R]^\omega$
as $\{D_{\alpha}:{\alpha}<2^{\omega}\}$.

By transfinite induction define
an increasing continuous sequence $(f_{\alpha}:{\alpha}\le 2^{\omega})$
of injective functions from subsets of $\mathbb R$ into $Y$ such that
$|f_{\alpha}|\le {\alpha}+{\omega}$, $D_{{\beta}}\subset dom(f_{\alpha})$
for ${\beta}<{\alpha}$, and
$$\text{ if ${\beta}<{\alpha}$ and }\overline D_{\beta}\ne \mathbb R
\text{ then }
\exists x_{\beta}\in (dom(f_{\alpha})\setminus \overline D_{\beta}) \
f_{\alpha}(x_{\beta})\in \overline{f_{\alpha}[D_{\beta}]}^\nu.
$$

Assume that ${\alpha}={\beta}+1$, and we have constructed $f_{\beta}$.
Let $g\supset f_{\beta}$ be an injective function from
$dom(f_{\beta})\cup D_{\beta}$ into $Y$.

If $\overline{D_{\beta}}=\mathbb R$, then$f_{\alpha}=g$ works.

Assume that $U=\mathbb R\setminus \overline{D_{\beta}}\ne \emptyset$.
Since $|U|=2^{\omega}$ and $|\overline{g[D_{\beta}]}^{\nu}|\ge 2^{\omega}$ we can pick $x_{\beta}\in U\setminus dom(g)$
and $y_{\beta}\in \overline{g[D_{\beta}]}^{\nu}\setminus ran(g)$.
Let
$$f_{\alpha}=g\cup\{(x_{\beta},y_{\beta})\}$$

This completes the inductive construction.
QED.

Pick a 0-dimensional $T_2$ space $(Y,\nu)$ which meets the requirements of the lemma.
(For example, $Y=\omega^*$ works)

Apply the lemma to obtain an injective
$f:\mathbb R\to Y$.

Define the topology $\tau$ by declaring that $f$
is a homeomorphism between $(\mathbb R,\tau)$ and $(f[\mathbb R],{\nu})$.

To show that $\tau $ is as required assume that $\emptyset\ne U\in \varepsilon $ such that $F=\mathbb R\setminus U$
is infinite. Let $D$ be a countable $\varepsilon$-dense subset of $F$.
Then there is $x_D\in \mathbb R\setminus \overline D$ such that

$f(x_D)\in\overline{f[D]}^{\nu}$.
Since $f$ is a homeomorphism, $x_D\in\overline{D}^{\tau}$.

Thus $D\subset F$ implies that $x_D\in \overline{F}^{\tau}\setminus F$,
and so $F$ is not $\tau$-closed. Thus $U\notin\tau$.

Thus we proved the proposition.

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