Imagine $\Bbb{R}^3$ with this equivalence relation $(x,y,z) \sim (x',y',z') \iff \exists t \in \Bbb{R}^3 - \{0\}: x'=tx, y'=ty, z'=tz$.
In this case how can I show that $\Bbb{R}^3/\sim$ is compact and hausdorff but it isn't orientable?
Thank you in advance!
This space is the same as $\mathbb R \mathbb P^2 \cup \{ 0 \}$.
But let us try to see it more directly. Let us call your space $X$ (and let us also forget the origin, since it just makes the computations messier). So $X= \mathbb R^3 - \{ 0\} / \sim $.
First, I claim that there is a map $\pi: \mathbb R^3 -\{0\} \to S^2$ and a map $\Pi:S^2 \to X$ such that the following diagram commutes: $$ \begin{array}{cccc} \mathbb R^3 - \{ 0\} & \xrightarrow{\pi} & S^2 \\ &\searrow & \downarrow \\ && X \end{array} $$
Indeed, $\pi$ is defined by $x \mapsto x/|x|$, and $\Pi$ is defined by sending $x$ to its equivalence class. Note that it is a $2:1$ map.
This proves that $X$ is compact, since it is the image of a compact space. It also proves Hausdorffness, because $\Pi$ is locally a diffeomorphism (explain why and why this is enough).
So $X$ can be identified with $S^2/\mathbb Z_2$, where $\mathbb Z/2$ acts by $x \mapsto -x$.
Here's a sketch of the non-orientability: the group action of $\mathbb Z / 2$ is not orientation-preserving on $S^2$, then use this old math.stachexchange answer.