A weaker Hausdorff topology on $\mathbb R$ with different system of compact subsets? Consider the real line $\mathbb R$ with the usual topology, generated by intervals $(a,b)\subseteq{\mathbb R}$.
Do there exist a weaker Hausdorff topology on $\mathbb R$ with different (a wider) system of compact subsets?
 A: In other words, you are asking for a continuous bijection from $\mathbb R$ to a Hausdorff space, which is not a homeomorphism. Let
$$S=\{(x,y)\in\mathbb R\times\mathbb R:(x\le1\wedge y=0)\vee(x^2+y^2=1\wedge y\gt0)\}.$$
Topologized as a subspace of $\mathbb R\times\mathbb R$, $S$ is is a Hausdorff space which is not homeomorphic to $\mathbb R$; and there is obviously a continuous bijection from $\mathbb R$ to $S$.
The weaker topology on $\mathbb R$ that you are looking for is one which makes $\mathbb R$ homeomorphic to $S$.
A: Let $\tau$ be the Euclidean topology on $\Bbb R$, and suppose that $\tau'\subseteq\tau$ is a Hausdorff topology on $\Bbb R$. Let $K$ be any $\tau$-compact subset of $\Bbb R$. Then the subspace topology on $K$ induced by $\tau$ is a compact Hausdorff topology, which means that it’s a minimal Hausdorff topology: no strictly weaker topology on $K$ is Hausdorff. Thus, $\tau'$ must induce the usual topology on $K$. In particular, $\tau'$ and $\tau$ must agree on each of the sets $[-n,n]$ for $n\in\Bbb Z^+$. 
Let $\Bbb R^*$ be the one-point compactification of $\Bbb R$, with topology $\tau^*$, and let $p$ be the point at infinity. Let
$$\tau'=\{U\in\tau:0\notin U\text{ or }U=V\cap\Bbb R\text{ for some }V\in\tau^*\text{ such that }0,p\in V\}\;.$$
This amounts to saying that every point of $\Bbb R\setminus\{0\}$ has its usual nbhds, and $\tau'$-open nbhds of $0$ are of the form $(\leftarrow,a)\cup U\cup(b,\to)$ for some $a,b\in\Bbb R$ and $\tau$-open nbhd $U$ of $0$. It’s easy to check that $\tau'$ is still Hausdorff. It’s also strictly weaker than $\tau$: for example, $(-1,1)\notin\tau'$, since it’s not a $\tau'$-nbhd of $0$. It turns out that this isn’t quite enough to pin down the topology, because a point can have nbhds that are locally Euclidean but have ‘tails’ far away from the point.
Let $K$ be any $\tau$-closed subset of $\Bbb R$ that contains $0$, and let $\mathscr{U}$ be a $\tau'$-open cover of $K$. There is a $U_0\in\mathscr{U}$ such that $0\in U_0$. But $\Bbb R\setminus U_0$ is a bounded, $\tau$-closed subset of $\Bbb R$, so it is compact in both topologies, and therefore $K$ is $\tau'$-compact. In particular, $[0,\to)$ is a $\tau'$-compact set that is not $\tau$-compact.
Added: It just occurs to me that $\langle\Bbb R,\tau'\rangle$ can also be described as the space that you get when you identify $0$ and $p$ in $\Bbb R^*$. If you take the two-point compactification of $\Bbb R$, you can identify both new points with $0$ to get my example, or you can identify the point at plus infinity with $0$ and throw away the other to get bof’s example, or you can identify one of the new points with $1$ and the other with $0$ to get two more examples (depending on which way round you do make the identifications).
