$T_2$ topologies that are "as disjoint as possible" Let $X$ be an infinite set. Are there Hausdorff topologies $\tau_1, \tau_2$ on $X$ such that $\tau_1\cap\tau_2 = \{\emptyset\} \cup \{U\subseteq X: X\setminus U\text{ is finite}\}$? 
(That is, the intersection is as small as it can get.) 
And what about the special case $X=\mathbb{R}$ and $\tau_1$ being the topology coming from the Euclidean metric?
 A: 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.
A: In the comments, Lajos Soukup pointed out the following paper, which seems to contain lots of relevant information:

D. Shakhmatov, M. Tkachenko, and R. G. Wilson, "Transversal and $T_1$-independent topologies," Houston Journal of Mathematics vol. 30, no. 2 (2004), pp. 421 - 433.

(I will give references to the paper below, but I will paraphrase the results to avoid defining all of their terminology.)
They answer Dominic's first question in the negative:
Proposition 3.2: If $X$ is countably infinite, then no two Hausdorff topologies on $X$ intersect in only the co-finite sets.
They are also able to strengthen this result under various extra hypotheses:
Theorem 3.3: It is consistent (it follows from $\neg$CH plus a fragment of MA) that if $|X| < \mathfrak{c}$ then no two Hausdorff topologies on $X$ intersect in only the co-finite sets.
Corollary 3.5: If $|X| < 2^{\aleph_1}$, then no two compact Hausdorff topologies on $X$ intersect in only the co-finite sets.
On the other hand, they show that Dominic's question can have a positive answer for some sets of larger cardinality:
Corollary 3.8: Let $|X| = 2^{2^{\aleph_0}}$. There are two topologies on $X$, both of which realize the compact Hausdorff space $\beta \omega - \omega$, such that the intersection of these two topologies is the set of co-finite subsets of $X$.
See also the preceding lemma, which is more general. (The proof seems to rely heavily on the fact that infinite closed sets have full cardinality, so the technique does not seem to apply to Dominic's second question.)
As for Dominic's second question, I can't seem to find an answer anywhere. But the authors of the above paper do still give us something:
Corollary to Proposition 3.1: If $\tau$ is a Hausdorff topology on $\mathbb{R}$ such that the only sets common to $\tau$ and the usual topology on $\mathbb{R}$ are the co-finite sets, then $\tau$ is countably compact and contains no non-trivial convergent sequences.
I was able to access the Shakhmatov-Tkachenko-Wilson paper here, and I also found some very interesting related information here, on some slides from a talk of Blaszczyk.
A: If $\tau_1$ and $\tau_2$ meet your requirement, then they are called $T_1$-independent. The following paper may contain useful information: Shakhmatov, D.; Tkachenko, M.; Wilson, R. G. Transversal and $T_1$-independent topologies. Houston J. Math. 30 (2004), no. 2, 421–433. MR2852951.
