Give an example of a topology that is not compact, not connected, and not Hausdorff here are my questions:
1) Give an example of a topology that is not compact, not connected, and not Hausdorff.
2) Give an example that is connected, but not compact and not Hausdorff.

This is the hint we have:
*i) Let $(X,\mathcal{T})$ and $(X',\mathcal{T}')$ be two topological spaces, and the collection
$$
\mathcal{B} = \bigl\{ \{0\} \times U \mid U \in \mathcal{T} \bigr\} \cup \bigl\{ \{1\} \times U' \mid U'\in \mathcal{T}' \bigr\}
$$
defines a basis for a topology $\mathcal{T}_\mathcal{B}$ on the disjoint union $X \cup X':= (\{0\} \times X) \cup (\{1\} \times X')$
*ii) If $X$ and $X'$ are not empty, the space $(X \cup X', \mathcal{T}_\mathcal{B})$ is not connected.
*iii) $(X \cup  X', \mathcal{T}_\mathcal{B})$ is Hausdorff if and only if $(X,\mathcal{T})$ and $(X',\mathcal{T}')$ are.
*iv) $(X \cup  X', \mathcal{T}_\mathcal{B})$ is compact if and only if $(X,\mathcal{T})$ and $(X',\mathcal{T}')$ are.

I think for 1), maybe we can let $(X,\mathcal{T})$ equal to some space that is not Hausdorff union $(X',\mathcal{T}')$ which is not compact. For instance, $X = (\{1,2,3\},\mathcal{T})$ which is not HSD, because the each compact set is closed, yet if we take $y = \{1\}$, and $y^c$ is not the same as $\mathcal{T}$, thus $y^c$ is not open. And let $X' = (3, \infty]$, because it's clearly not compact. And $X$ and $X'$ are not empty, so it's not connected. Will this work?
For 2), let $X = (\{1,2,3\},\mathcal{T})$, and let it union with some set? 
Please give some example and give some explanation for why it is a good example.
Thank you for the help!
 A: Consider taking $\mathcal{T}_{fin} \sqcup \mathcal{T}_{uplim}$ on $[0,1]$. Since ony one is compact, and the other is Hausdorff, then the disjoint union will neither be connected, compact, and Hausdorff as you wish.
A: Good Morning Captain already gave the answer for the first question.
Here is an answer for the second question:

2) Give an example that is connected, but not compact and not Hausdorff.

Let $X=[0,1]\subset\mathbb{R}$ with the finite complement topology $\mathcal{T}_\mathrm{fin}$, and $Y=(0,1)\subset\mathbb{R}$ with the standard topology $\mathcal{T}_\mathrm{st}$.
We know that (or it is easy to show)


*

*$X$ is compact, connected, but not Hausdorff.

*$Y$ is connected, Hausdorff, but not compact.


Then $X\times Y$ with the product topology is connected but neither compact nor Hausdorff. The reason is as follows:


*

*$X\times Y$ is connected since both $X$ and $Y$ are connected.

*$X\times Y$ is not compact. If so, the image of the projection map $\pi_2\colon X\times Y\to Y$ should be compact, which contradicts $Y$ is not compact.

*$X\times Y$ is not Hausdorff because it has a homeomorphic copy of $X$ as a subspace which is not Hausdorff.


[Update] Similarly we could get the answer for the first question. Borrowing Good Morning Captain's idea:

1) Give an example of a topology that is not compact, not connected, and not Hausdorff.

Let $X=[0,1]\subset\mathbb{R}$ with the finite complement topology $\mathcal{T}_\mathrm{fin}$, and $Y=[0,1]\subset\mathbb{R}$ with the upper limit topology $\mathcal{T}_\mathrm{uplim}$.
We know that


*

*$X$ is compact, connected, but not Hausdorff.

*$Y$ is not compact, not connected, but Hausdorff.


Then $X\times Y$ with the product topology is not connected, not compact, not Hausdorff. The reason is the same as above.
A: *

*N with the base { {1,2}, {n} : n in N, n > 2 }.  

*N with the base { {1,n} : n in N }.  


Details left to the diligent reader.
