Suppose $X$ is Hausdorff. The family of continuous functions $f_\alpha:X\rightarrow[0,1]$ (indexed by $\alpha \in$ some given set $I$) has the property: if $x\in X$ and $A$ is closed subset of $X$ not containing $x$, then there exists an $\alpha \in I$ such that $f_\alpha(x)=0$ and $f_\alpha(A)=\{1\}$. The question is to show that the map $f=\Pi f_\alpha:X\rightarrow [0,1]^I$ is a topological embedding, where $[0,1]^I$ is given the product topology.
$f$ is of course continuous. $f$ is injective since by taking $A$ above to be a point $y$ (which is closed in $X$ by Hausdorff) different from $x$ we see that $f(x)$ is different from $f(y)$ at some coordinate.
But to show it is open (or closed) is beyond me. Please help! Thank you very much.
HINT: Let $U$ be a non-empty open set in $X$. For each $x\in U$ there is an $\alpha(x)\in I$ such that $f_{\alpha(x)}(x)=0$ and $f_{\alpha(x)}[X\setminus U]=\{1\}$; let $V_x=\{p\in f[X]:p_\alpha\ne 1\}$. Show that $V_x$ is open in $f[X]$ and that $f[U]=\bigcup_{x\in U}V_x$.