Explicit description of a topology that proves $\mathbb{R}$ is not minimal Hausdorff. Let $X$ denote a set, and $\tau \subseteq \mathcal{P}(X)$ denote a topology. Then $\tau$ is called minimal Hausdorff iff every strictly smaller subset of $\mathcal{P}(X)$ fails to be a Hausdorff topology. It turns out that every compact Hausdorff topology is minimal Hausdorff, so that in particular the affinely extended real line $[-\infty,\infty]_\mathbb{R}$ is compact Hausdorff.
A suggestion was made here that $\mathbb{R}$ is not minimal Hausdorff. Quoting Andreas Blass:

I don't think $\mathbb{R}$ works. There's a coarser Hausdorff topology obtained
  by making everything that used to diverge to $+∞$ converge to $0$ instead.

Explicitly, what is this topology?
 A: Note: A previous version of this answer, though giving a correct example of a weaker topology that is Hausdorff, didn't correctly describe the topology mentioned by the OP. 
It's a P-shape.
Define a bijection from $\mathbb{R}$ to the P as follows. You start at the bottom when $t = -\infty$, move up the side, go through the intersection point when $t = 0$, and then wind around so that you come back to the intersection point at $t = +\infty$.
Now import the topology of the P to $\mathbb{R}$ via the bijection just defined. The topology is strictly weaker than the ordinary topology because the mapping from $\mathbb{R}$ to the P is continuous, but not a homeomorphism.
A: Keep reading the comment:

(Pictorially, take the $+\infty$ end of $\mathbb R$ and bend it back to approach 0.)  

If you don't get that, think of it this way:
Consider $S^1$, embedded in $\mathbb R^2$, as a one point compactification of the real line. Now pinch the points corresponding to $0$ and the additional point together. This gives you a figure 8 in the plane. Give it the subspace topology, and compare.
The new space is the same as the real line, except it effectively lost neighborhoods of 0. Every neighborhood of 0 now contains infinite open intervals on both sides. It is missing those of the form $(-a,b)$, $0<a,b<\infty.$ Therefore the topology is coarser.
Edit: Mike's $P$ interpretation is probably the more accurate one of the specific comment in question. Both demonstrate a coarser topology fairly explicitly, as each loses neighborhoods of zero by forcing them all to have infinite intervals.
A: Here is an explicit description in terms of nbhds. If $x\ne 0$, a set $A\subseteq\Bbb R$ is a nbhd of $x$ if and only if there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq A$. A set $A$ is a nbhd of $0$ if and only if there are an $\epsilon>0$ and an $n\in\Bbb Z^+$ such that $(x-\epsilon,x+\epsilon)\cup(n,\to)\subseteq A$. (Note that a nbhd need not be open: a set is open if it is a nbhd of each of its points.)
