Proving that a function between two specific topological spaces can't be both continuous and injective I'm having trouble solving this problem.
Let $\mathcal{B}=\{\emptyset,(m,n)\}_{m,n\in\mathbb{Z}}$ a family of subsets, prove it is a base for $\mathbb{R}$, and let $Y=(\mathbb{R},\mathcal{U}_B)$. Then consider the following topological space:
$$X=\mathbb{R}\times Y $$
where $\mathbb{R}$ has the usual topology.
The problem is:
prove that there are no function $f:X\to\mathbb{R}^2$ that are continuous and injective.
Regarding the first part, I showed that $\mathcal{B}$ is a base for $\mathbb{R}$.
Coming to the second part, I can't prove there are no such function.
I tried assuming $f$ is injective and then proving it can't be continuoius (and viceversa), but I am stuck.
Assuming $f$ is injective, I considered an open set $A\subseteq \mathbb{R}^2$ and considered $f^{-1}(A)\subseteq X$. I want to show $f^{-1}(A)$ is not open in $X$ so it cannot be seen as $f^{-1}(A)=U\times V$ where $U,V$ are open sets of respectively  $\mathbb{R}$ and $Y$.
Does anyone have an idea about how to procede?
 A: Here is the solution I came up with.
Since $X$ is not Hausdorff, exist at least two different point that are not separable. We name this two points $(x_1,y_1)$ and $(x_2,y_2)$. Let $f:X\to \mathbb{R}^2$ injective, it follows that $f(x_1,y_1)\neq f(x_2,y_2)$. But $\mathbb{R}^2$ is Hausdorff, then exists open sets $V=U_{f(x_1,y_1)}$ and $W=U_{f(x_2,y_2)}$ s.t $f(x_1,y_1)\in V$,$f(x_2,y_2)\in W$ and $V \cap W=\emptyset $. If $f$ is continuous, $f^{-1}(V)$ and $f^{-1}(W)$ are open and in particular $f^{-1}(V) \cap f^{-1}(W)=\emptyset$ and hence $(x_1,y_1)$ and $(x_2,y_2)$ are separable. Whence $f$ can't be continuous.
Conversely, let $f$ be a continuous function. If we assume $f$ as injective,
we have the following condition: $(x_1,y_1)\neq(x_2,y_2) \implies f(x_1,y_1)\neq f(x_2,y_2)$. Since $f(x_i,y_i)$ for $i=1,2$ is in $\mathbb{R}^2$, and is $\mathbb{R}^2$ Hausdorff there, must exist open sets $V=U_{f(x_1,y_1)}$ and $W=U_{f(x_2,y_2)}$ s.t $f(x_1,y_1)\in V$,$f(x_2,y_2)\in W$ and $V \cap W=\emptyset $. Now, since $f$ is continuous, $f^{-1}(V)$ and $f^{-1}(W)$ are open in $X$ and $f^{-1}(V) \cap f^{-1}(W)=\emptyset$. Applying this reasoning, we obtain that all points are separable in $X$, that is absurd since $X$ is not Hausdorff. Whence $f$ can't be injective.
Is this acceptable?
