Topology on $\mathbb{R}$ strictly coarser (resp. finer) than the usual one which is still Hausdorff (resp. connected) The following are simple observations.

Suppose $\mathcal{T}_1,\mathcal{T}_2$ are two topologies on a set $X$ such that $\mathcal{T}_1$ is finer than $\mathcal{T}_2$.

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*If $( X ,\mathcal{T}_2 )$ is Hausdorff, then so is $( X ,\mathcal{T}_1 )$.

*If $( X ,\mathcal{T}_1 )$ is connected,  then so is $( X ,\mathcal{T}_2 )$.


The real numbers with the usual topology is both Hausdorff and connected, which leads to the following two questions.

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*Is there a topology on the real numbers strictly coarser than the usual one, which makes it still Hausdorff?


*Is there a topology on the real numbers strictly finer than the usual one, which makes it still a connected space?
 A: People suggesting that you can't make a coarser, still Hausdorff topology are wrong. Their proofs are invalid, and so is the conclusion. I construct a coarser topology that is still Hausdorff. You can think of this space as adding a point at $+\infty$ and gluing it to 0.
Let $\tau$ be the usual topology of open sets on $\mathbb{R}$. Define $\tau' \subset \tau$ as the collection of sets in $\tau$ that, if they contain 0, contain some interval $(\alpha,+\infty)$. That is, the open sets in our new topology are the old open sets, but, if they contain the point 0, we also require them to be a neighborhood of $+\infty$. By definition $\tau'$ is coarser than $\tau$, and it is immediate that it is closed under finite intersection and arbitrary union. To show it's Hausdorff, say we're trying to separate two points $x$ and $y$. If $x,y \neq 0$, just choose some small intervals around $x$ and $y$, making sure you don't include 0 in either of them. If one of them is zero, just pick intervals as usual, and throw in some $(\alpha,+\infty)$ to the neighborhood of 0, with $\alpha$ large enough not to include the other point.
Someone already provided a working example for a finer topology that's still connected.
Edit: I'll provide some informal rationale about why the compactness of closed intervals is not enough to get what people seem to want, and shouldn't be. Morally, you can think of Hausdorff as meaning "sequences can't converge to more than one point", and a coarser topology as being one in which "more converging happens." The reason that you can't make a coarser Hausdorff topology on a compact space is that there is already lots of converging going on. To get more converging, you would have to make a sequence converge to something new, when it already converges to something else (breaking Hausdorff). The key way we're using compactness here is that we know that "everything that could converge already does"; that's a global property of compactness, and we can't recover it from local compactness, or knowing that our space is a countable union of compact sets, or anything like that. In the construction I gave, we added "more convergence" by saying that sequences that tend to $+\infty$, which ordinarily diverge, should instead be considered converging to 0.
A: For your second question, yes.
Let $\tau$ be the topology on $\mathbb R$ generated by $X\cup \mathbb Q$ or $X$, where $X$ is open in $\mathbb R$ in the usual topology. Then it is connected.
To see this, let $\mathbb R = Y_1 \cup Y_2$ be two disjoint union, where $Y_i \in \tau$. Then either one of them, say $Y_1$, is not open in the usual topology. By definition,
$$Y_1 = X_1 \cap (W_1 \cup \mathbb Q),$$
where $X_1, W_1$ are open (in the usual sense). As this is not open in the usual topology, there is $p \in Y_1\cap \mathbb Q$ such that $Y_1$ does not contain any open interval around $p$, but $Y_1$ does contain all rational numbers sufficiently close to $p$.  But then it is impossible for $Y_2$ to be open in $\tau$: $Y_2$ must contain irrational numbers arbitrarily close to $p$, but then openness would force it to also contain rational numbers arbitrarily close to $p$.
A: In addition to the counterexamples, the true statement should be that for a compact Hausdorff space, you cannot weaken it without losing the $T_2$ axiom and you cannot strengthen it without losing the compactness (see Functional Analysis by Rudin). To see that, suppose $\tau_1\subseteq\tau_2$ are two topologies defined on the same space such that $\tau_1$ is Hausdorff and $\tau_2$ is compact. If $F$ is a $\tau_2$-closed set, then it is compact as is in a compact space. Then any $\tau_1$-open cover of $F$ is automatically a $\tau_2$-open cover. Since $\tau_2$ is compact, there should be a finite sub-cover. Thus $F$ is a $\tau_1$ compact set. Note that $\tau_1$ is Hausdorff. This means $F$ is $\tau_1$-closed. In a word, a $\tau_2$-closed set is $\tau_1$-closed, showing that the two topologies are actually equal. 
