This answer says the following:
The converse need not be true: Take $X=I\cup\{0'\}$, a unit interval with two origins, where the neighborhood base of $0'$ is formed by intervals $[0',\epsilon):=\{0'\}\cup(0,\epsilon)$.
Where exactly is $0'$ on the line? From the rest of the answer, it seems like we require $0' \in \mathbb{R}\setminus I$, but I'm honestly not sure why it deserves to
Let $I=[0,1]$. This construction is precisely the quotient space $$X=(I\times \{0\}\sqcup I\times \{1\})/{\sim}$$ where $(x,1)\sim (x,0)$ if and only if $x\ne 0$. Then this quotient space "looks" like a line with two origins: namely $(0,1)$ and $(0,0)$, or $0'$ and $0$ if you wish. More intuitively, we take the disjoint union of two unit intervals, and then we "glue" all of the corresponding points together except for the origins. In this way, we get a "unit interval" with two origins.
In general, when we make these sorts of "quotient" constructions, we define the open sets to be precisely those sets that make the quotient map continuous. In particular, $U\subset X$ is open if and only if $\pi^{-1}(U)$ is open in $I\times \{0\}\sqcup I\times \{1\}$. Here $\pi: (I\times \{0\}\sqcup I\times \{1\})\to X$ is the map which sends a point to its equivalence class.
So, this lets us see that the open sets in this line with two origins containing the point $0'$ are precisely those of the form $[0', k)$ for $k\in (0,1)$. Indeed, then, a neighborhood basis for $0'$ should look like a neighborhood basis for $0:$ i.e., it is comprised of open sets of the form $[0',\epsilon)$ with $\epsilon\in (0,1)$.
$0'$ isn't on the line at all. It's an artificial element.
Think of it like this: Imagine two copies of $I$, arranged parallel. "Zip" them together, starting from the ends; partway along, we have a sort of almond shape with little lines at either end.
Now imagine we keep going, until the only points not zipped together are the origins of each interval. Neighborhoods of each origin bleed into the zipped part - so, for example, $\frac{1}{4}$ is "close" to both $0$ and $0'$. To get to $0'$, move toward $0$ and take a left at the last moment.
