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!

*

*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.
