I'm reading a book, where there is the following theorem: Let $X$ be a topological Hausdorff space. Then the statements are equivalent:
(a) $X$ is completely regular.
(b) A net $\{x_i\}$ in $X$ converges to $x$ if and only if $f(x_i) \to f(x)$ for each continuous function $f: X \to [0,1]$.
From the proof in the book it is clear, that $(a) \implies (b)$, but i think the direction $(b) \implies (a)$ is left out or i just don't get the argument. Could someone please help me to understand this?
If $X$ is not completely regular, then there is a point $x$ and an open neighborhood $U$ of $x$, such that for any continuous map $f:X\to[0,1]$, no open neighborhood $V$ around $f(x)$ has its preimage in $U$, or, in other words, the closed set $A:=X\setminus U$ has points mapped to $V$ for any $V$ around $f(x)$.
Now we can construct a net $\Phi:\{V\text{ open}\mid f(x)\in V\}\to X $ such that $f(Φ)\to f(x)$ but $Φ\not\to x$, by choosing for each neighborhood $V$ around $f(x)$ a point $Φ(V)\in A\cap f^{-1}(V)$.