Topological space that is compact and $T_1$ but not Hausdorff (i.e. normal but not hausdorff) I was freshing up on some topology, and this text I'm reading mentions T1 does not imply Hausdorff. A few counter-examples are readily available, like the natural numbers under the co-finite topology. 
But what if we place a restriction on the space to also be compact, the text doesn't mention anything about that and I can't come up with any examples of spaces that are compact and T1 but not Hausdorff. To state it otherwise, I'm looking for a space that is T1 but not normal(=compact and Hausdorff). 
 A: You've already given an example: the natural numbers (or any infinite set, really) under the co-finite topology. Given any open cover, fixing a single (non-empty) element of the cover yields an open set that has all but finitely-many of the natural numbers as elements. Thus, only finitely-many more elements of the cover are needed, forming a finite subcover.
A: The affine space $\Bbb C^n$ endowed with the Zariski topology: $C$ is closed if and only if there is a family of multivariate polynomials $\mathcal F_C\subseteq \Bbb C[x_1,\cdots,x_n]$ such that $$C=\{x\in\Bbb C^n\,:\,\forall f\in\mathcal F_C,\ f(x)=0\}$$
These topologies have the following notable properties:


*

*they are T1;

*any two non-empty open sets have non-empty intersection (irreducibility): thus the topology is not T2, and on a side note all open subsets are connected;

*for any non empty family $\mathfrak C$ of closed sets, there is $C\in\mathfrak C$ which is maximal with respect to inclusion "$\subseteq$" (noetherianity): thus every subspace is compact.


The proofs of these facts are not difficult, but they use a couple of lemmas of commutative algebra which might make the exposition a bit long.
An easier special case of this is when $n=1$, in which case the Zariski topology is just the cofinite topology on $\Bbb C$ (the topology where a set is open if and only if its complement is either finite or the whole space).
Addendum: I assumed here that your definition of "compact topological space" is:

Cpt: A topological space such that any open cover $U$ admits a finite subcover.

However, a considerable number of authors (for instance, Bourbaki), call "quasi-compact" a topological space which satisfies Cpt and "compact" a T2 quasi-compact space. In this case, though, your question would be trivial.
A: Let $T_X$ be a compact Hausdorff topology on an infinite set $X$,  with no isolated points. So every non-empty member of $T_X$ is an infinite set. Let $Y=X\times \{1,2\}.$ $$\text {Let} \quad B= \{\;( t\times \{1,2\} )\; \backslash A:t\in T_X \land A \text { is finite} \}.$$ Then $B$ is a base for a topology $T_Y$ on $Y.$ 
(i). For each $p\in Y$ we have $Y$ \ $\{p\}=(X\times \{1,2\})$ \ $\{p\}\in B\subset T_Y.$  So $T_Y$ is a $T_1$ topology.
(ii). For $x\in X$ and $i\in \{1,2\},$ if $(x,i)\in U_i\in T_Y,$ then $(x,i)\in (t_i\times \{1,2\})$ \ $A_i\subset U_i$ for some $t_i\in T_X$ and some finite $A_i.$  
Then $U_1\cap U_2\supset (t_1\cap t_2)$ \ $(A_1\cup A_2)\ne \phi .$ So $(x,1)$ and $(x,2)$ do not have disjoint nbhds: $T_Y$ is not a Hausdorff topology.
(iii). For $i\in \{1,2\}$ the subspace $X\times \{i\} $ is homeomorphic to $X$ so it is compact. Now $Y$ is the union of the two compact subspaces $X\times \{1\},\;X\times \{2\}$ so $Y$ is also compact.
