Construction of a non-Hausdorff space. The following is a formulation for the construction of a non-Hausdorff space.

Let $X = \mathbb{R}$ with the standard topology. Let $f,g$ be functions from $X$ to $\mathbb{R}^2$ such that, for every $x \in X$, $f(x) = (x,0)$ when $x \neq 0$ and $f(x) = (0,1)$ when $x = 0$. Similarly, $g(x) = (x,0)$ when $x \neq 0$ and $g(x) = (0,2)$ when $x = 0$. Let the union of the image sets of $f$ and $g$, call it $Y$, have the finest topology respect to which both the maps are continuous.  Then $Y$ is a not a Hausdorff space.

Over here, $Y$ is just the x-axis with the two extra points $\{(0,1),(0,2)\}$. The problem is I am unable to see why this space is not Hausdorff. All I have been able to show is that the set $$ \{ x \in \mathbb{R} \mid f(x) = g(x)\}$$ is not closed in $X$, hence $Y$ is not Hausdorff. 
But I am not convinced I am correct. Could someone please comment on the validity of the process ? If instead of 2, I had $n$ functions would the space still be non-Hausdorff ?
 A: First, to construct a non-Hausdorff space doesn't necessitate so involved an example, since the trivial topology on a set of size $>1$ gives such a space.
However, in this case $Y$ isn't Hausdorff because, if $(0,1)\in U$, $(0,2)\in V$, and $U\cap V=\emptyset$ are both open, then $f^{-1}(U)$ and $g^{-1}(V)$ are both open and $f^{-1}(U)\cap g^{-1}(V)=\{0\}$. Thus $\{0\}\subseteq\mathbb{R}$ is also open, in contradiction.
A: It is true that $\{x\in\mathbb R:f(x)=g(x)\}=\mathbb R\setminus \{0\}$ is not closed in $X$, but why does this imply that $Y$ is not Hausdorff?
One way to demonstrate the fact that $Y$ is not Hausdorff is to show that if $U$ and $V$ are open neighborhoods of $(0,1)$ and $(0,2)$ respectively, then $U\cap V\neq\emptyset$.
Because $Y$ is equipped with the final topology with respect to $f$ and $g$, we know $f$ is continuous and so $f^{-1}(U)$ is open in $\mathbb R$. In particular $0\in f^{-1}(U)$ since $f(0)=(0,1)\in U$, so there exists some $\epsilon_1>0$ such that $(-\epsilon_1,\epsilon_1)\subset f^{-1}(U)$. Thus $$f((-\epsilon_1,\epsilon_1))=\{(x,0):0<|x|<\epsilon_1\}\cup\{(0,1)\}\subset U.$$
Similarly, $g$ is continuous and so $g^{-1}(V)$ is open in $\mathbb R$. In particular $0\in g^{-1}(V)$ since $g(0)=(0,2)\in V$, so there exists some $\epsilon_2>0$ such that $(-\epsilon_2,\epsilon_2)\subset f^{-1}(V)$. Thus
$$
g((-\epsilon_2,\epsilon_2))=\{(x,0):0<|x|<\epsilon_2\}\cup\{(0,2)\}\subset V.
$$
Take $\epsilon=\frac{1}{2}\min(\epsilon_1,\epsilon_2)$. Then $(\epsilon,0)\in U\cap V$ so $U\cap V\neq\emptyset$.
