Questions regarding a equivalencerelation on X. $X=[-1,1]\times \mathbb{Z}$ and deﬁne an equivalence relation on X via
$(t,n)$~$(t´,n´)$ if and only if $(t,n)$=$(t´,n´)$ or $t=t´\ne 0$.
(You may assume that this is indeed an equivalence relation.) Let $p: X → X/∼$
denote the quotient map and set Y := X/∼.
a) Show that Y is not compact
b) Show that Y is not Hausdorff
c) Show that $p([−1, 1]\times  \{0\})$ is compact and calculate its closure.
a) and b) I'm not really sure how to proceed)
But in c) I thought that the set $[−1, 1]\times  \{0\}$ is a product of two compact spaces hence compact. And the image of a compact space under a continuous map is compact.
The image $p([−1, 1]\times  \{0\})$ is the equivalence class $[(t,0)]$, but I'm quite unsure what it even means to be open in this space X~.
 A: Notice that $[(0,n)]=\left\{(0,n)\right\}$. Clearly this element is not open in $Y$ as $p^{-1}([(0,n)])=\left\{(0,n)\right\}$ is not open in $X$. This observation is going to be fundamental in understanding this problem.
If we want to find an open cover of $Y$ that doesn't admit a finite subcover, we should focus on the points $[(0,n)]$. Let's find opens in $Y$ that actually contain the element $[(0,n)]$. 
Consider $O_n:=p^{-1}(\bigcup_{\alpha\in [-1,0)} [\alpha,0]\cup [0,n]\cup \bigcup_{\beta\in (0,1]}[\beta,0])$. By drawing this, is not difficult to see that this set is open in $X$. Hence $Y_n:=\bigcup_{\alpha\in [-1,0)} [\alpha,0]\cup [0,n]\cup \bigcup_{\beta\in (0,1]}[\beta,0])$ is open in $Y$ and moreover $[0,m]\in Y_n$ if and only if $n=m$.
Notice that $\bigcup_{n\geq 0}O_n=Y$ but it does not admit a finite subcover since we need all $O_n$ to cover the elements $[0,m]$. This shows 1). Try to understand this argument (by drawing it), maybe this helps you to find the other yourself. Good luck.
