The Line with two origins I have seen descriptions of the "line with two origins" using quotient spaces. My professor has defined it in an alternate way. However, I can not wrap my head around how the following descriptions forms a line with two origins. 
Consider $X=\mathbb{R} \setminus \{0\} \cup \{p,q\}$, that is $X$ is the union of the reals minus $0$, and two points. Consider sets of the type
$U_a = (-a,0) \cup \{p\} \cup (0,a)$
$V_a=(-a,0) \cup \{q\} \cup (0,a)$
where $a >0$. And let
$\mathcal{B}=\{U_a\}_{a>0} \cup \{V_a\}_{a >0} \cup \{ \text{all open intervals of} \hspace{2mm} \mathbb{R} \hspace{2mm} \text{not containing the origin} \}$
Then $\tau=\{\bigcup_{\alpha} B_{\alpha} \big | B_{\alpha} \in \mathcal{B} \}$.
How is this a line with two origins?
 A: Let's make the construction a little simpler.
We start with a line, $\Bbb R$.

Then we removed the origin $0$, so $\Bbb R\setminus\{0\}$.  Let's call that $\def\rstar{\mathbb R^\star}\rstar$.

Then we add a new point $p$—a new point, not a real number—and we have $\Bbb \rstar\cup\{p\}$.
And the way we add back this point is special.  Where before we had some open set $U$ that contained $0$, we now have an open set $U\setminus\{0\}\cup\{p\}$.  This set is exactly $U$, but with $0$ replaced by $p$. Where before we had an open set $V$ not containing $0$, we keep $V$ unchanged; $V$ is still open.

So the open sets are just like the ones we had before, except that $0$ has been replaced by $p$.  Topologically, the point $p$ behaves just like $0$ did before.  Where before we had $0$ in some open set $U$, we now have $p$ in some analogous open set $U\setminus\{0\}\cup\{p\}$. $p$ is a perfect replacement for the origin $0$ that we deleted.  Really it's just $0$, but with another name.
This new space, $\Bbb R\setminus\{0\}\cup\{p\}$, is exactly $\Bbb R$, except that $0$ has been removed and replaced  by $p$.  It's easy to show that this space is topologically identical to $\Bbb R$.  The homeomorphism is particularly simple: it is the identity function, except that it takes $0$ to $p$, because $\Bbb R\setminus\{0\}\cup\{p\}$ has  has $p$ instead of $0$.
Got that?

Now we add another new point $q$, in exactly the same way we added $p$: if $U$ was open before, and $0\in U$, then $U\setminus\{0\}\cup\{q\}$ is open now.  This set is exactly $U$, but with $0$ replaced by $q$.

So now we have something like $\Bbb R$, except it has this extra point $q$.  But $q$ has all the same properties that $p$ has!  And in particular, just as $p$ was a perfect replacement for the origin $0$ that we deleted, $q$ is also a perfect replacement for the origin $0$ that we deleted.
And just as $\rstar\cup\{p\}$ was homeomorphic to $\Bbb R$, so is  $\rstar\cup\{q\}$ homeomorphic to $\Bbb R$.
But this new space is $\rstar\cup\{p,q\}$ and has both $p$ and $q$. We can delete either one of them and get a space identical to $\Bbb R$.
We deleted the origin $0$ and replaced it with $p$ and with $q$, so we now have something like $\Bbb R$, except that instead of one origin it has two, $p$ and $q$.

So it is called the line with two origins.
