Today we did this proof, but we could not finish it and our prof said that the end would be easy, but I could not finish this proof.
Let $X$ be a $T_3$ space with a countable basis $B$. Then we showed that $\forall x \in X \forall V_2 \in B$ with $x \in V$, there is a set $V_1 \in B$, such that $(V_1,V_2) \in P:=\{(V_1,V_2) \in B \times B: \exists f:X \rightarrow [0,1] \text{ continuous}: f(V_1) \subset [0,1/2), \text{ and } f|_{V_2^C} =1\}\text{.}$ This part was alright. So we said that we would get a countable number of these continuous maps (as you get one for every element and for every open set of the basis containing this element) and by putting them together to a map $F:X \rightarrow H$, where $H$ denotes the Hilbert cube such that $x \rightarrow (f_1(x),f_2(x),...)$, we just need to show that this one is injective and the inverse also continuous.
We said that we should fix a $x_0 \in X$ and a closed set $A \subset X$ with $x_0 \notin A$. Then $A^C$ is open and contains $x_0$. Then there is a open set $V_2$ of the basis such that $x_0 \in V_2 \subset A^C$ (because $A^C$ is open it is the union of basis sets). Now by the same reasoning as above we find a $V_1$ such that $(V_1,V_2) \in P$.
Hence, there is a corresponding continuous map such that $f_(x_0) < \frac{1}{2}$ and $f|_{V_2^C}=1$. Therefore $f(x_0) \notin \overline{f(A)}$.And if there is a component that does not do it, then we also have $F(x_0) \notin \overline{F(A)}$.
Hence, the map is injective cause we can take $A:=\{x_1\}$, with $x_1 \neq x_0$ and then $f(x_0) \neq f(x_1)$ and therefore also $F(x_0) \neq F(x_1)$, but why is the inverse of $F$ continuous?
Let $A \subset X$ be closed. To show continuity of $G:= F^{-1}: F[X] \rightarrow X$ (which by injectivity is well-defined!), it suffices to show that $G^{-1}[A]$ is closed in $F[X]$. Note that $G^{-1}[A] = F[A]$. So we need to show that $F[A]$ is closed in $F[X]$, and this is what follows from the argument:
Let $y \in F[X]$, so $y = F(x_0)$ for some $x_0 \in X$, be in $\overline{F[A]}$, then the argument as given shows that $x_0 \in A$, as $x_0 \notin A$ leads to $F(x_0) \notin \overline{F[A]}$, as seen. But then $y = F(x_0) \in F[A]$, showing $F[A] = G^{-1}[A]$ to be closed, as required.