Eric van Douwen’s paper ‘A regular space on which every continuous real-valued function is constant’, Nieuw Arch. Wisk. $30$ $(1972)$, $143$-$145$, actually gives a ‘machine’ for starting with a $T_3$ space having two points that cannot be separated by a continuous real-valued function and producing from it a $T_3$ space on which all continuous real-valued functions are constant. I no longer have a copy of the paper or easy access to it, so I’m working from memory, but the construction below is either the same as or very similar to Eric’s.
Let $Z$ be any $T_3$ space with points $p$ and $q$ that cannot be separated by a continuous real-valued function. Let $Y=Z\setminus\{p,q\}$, and let $\kappa=|Y|$. Give $\kappa$ the discrete topology, and let $X=\kappa\times Z$ with the resulting product topology. For notational convenience let let $X_\xi=\{\xi\}\times Z$, $p_\xi=\langle\xi,p\rangle$, and $q_\xi=\langle\xi,q\rangle$ for $\xi<\kappa$. Clearly $\kappa$ is infinite, so there is a bijection $\varphi:\kappa\to\kappa\times Y\subseteq X$. We may (and do) further assume that if $\varphi(\xi)=\langle\eta,y\rangle$, then $\eta\ne\xi$.
Define an equivalence relation $\sim$ on $X$ as follows.
- Of course $x\sim x$ for all $x\in X$.
- We identify all copies of $p$ to a single point: $p_\xi\sim p_\eta$ for all $\xi,\eta<\kappa$.
- For all $\xi<\kappa$, $q_\xi\sim\varphi(\xi)$; this identifies each point of $\kappa\times Y$ with a unique $q_\xi$.
Let $X'=X/\!\sim$; we hope to show that $X'$ is $T_3$, and that every real-valued continuous function on $X'$ is constant.
Think of $Z$ as a string with endpoints $p$ and $q$. The idea of $\sim$ is to glue all of the copies of $p$ together into one point $p^*$, and then to run a separate string from $p^*$ to each other point, gluing the $q$ end of that string to the point. (I’ve always thought of this construction as Eric’s Spaghetti Machine.) This ought to ensure that if $x$ and $y$ are any points of $X'$, and $f:X'\to\Bbb R$ is continuous, the string from $p^*$ to $x$ ensures that $f(x)=f(p^*)$, while the string from $p^*$ to $y$ ensures that $f(y)=f(p^*)$, so that $f(x)=f(y)$.
And in fact it does. Let $\pi:X\to X'$ be the canonical quotient map, and let $p^*=\pi(p_0)$ (which is of course the same as $\pi(p_\xi)$ for each $\xi<\kappa$). If $f:X'\to\Bbb R$ is continuous, then $f\circ\pi:X\to\Bbb R$ is continuous. Let $g=f\circ\pi$; then $g(p_\xi)=g(q_\xi)$ for all $\xi<\kappa$, since $g\upharpoonright X_\xi$ is a continuous real-valued function on $X_\xi$, which is homeomorphic to $Z$. Let $x'\in X'$ be arbitrary. Then $x'=\pi\big(\langle\eta,y\rangle\big)$ for some $\langle\eta,y\rangle\in\kappa\times Y$, and there is a $\xi\in\kappa\setminus\{\eta\}$ such that $\varphi(\xi)=\langle\eta,y\rangle$. Then $\pi(q_\xi)=\pi\big(\varphi(\xi)\big)=\pi\big(\langle\eta,y\rangle\big)$, so $f(x')=g\big(\langle\eta,y\rangle\big)=g(q_\xi)=g(p_\xi)=f(p^*)$, and $f$ is indeed constant on $X'$.
It’s actually harder to show that $X'$ is $T_3$.
Let $x'\in X'\setminus\{p^*\}$, and let $U$ be an open nbhd of $x'$. There is some $\langle\eta,y\rangle\in\kappa\times Y$ such that $x'=\pi\big(\langle\eta,y\rangle\big)$. Let $V_0=X_\eta\cap\pi^{-1}[U]$; $V_0$ is an open nbhd of $y$ in $X$. Let
$$K_0=\{\xi<\kappa:\varphi(\xi)\in V_0\}=\{\xi<\kappa:\pi(q_\xi)\in\pi[V_0]\}\;,$$
and let $$V_1=\bigcup_{\xi\in K_0}\big(X_\xi\cap\pi^{-1}[U]\big)\;.$$
In general, given $V_n$ for some $n\in\omega$, let
$$K_n=\{\xi<\kappa:\varphi(\xi)\in V_n\}=\{\xi<\kappa:\pi(q_\xi)\in\pi[V_n]\}\;,$$
and let $$V_{n+1}=\bigcup_{\xi\in K_n}\big(X_\xi\cap\pi^{-1}[U]\big)\;.$$
Finally, let $V=\bigcup_{n\in\omega}V_n$; then $V=\pi^{-1}[U]$.
There is an open nbhd $W_0$ of $\langle\eta,y\rangle$ in $X$ such that $\operatorname{cl}_XW_0\subseteq V_0$. For each $\xi\in\bigcup_{n\in\omega}K_n$ there is an open nbhd $G_\xi$ of $q_\xi$ in $X$ such that $G_\xi\subseteq X_\xi$, and $\operatorname{cl}_XG_\xi\subseteq\pi^{-1}[U]$. Given $W_n$ for some $n\in\omega$, let $L_n=\{\xi<\kappa:\pi(q_\xi)\in\pi[W_n]\}$, and let
$$W_{n+1}=W_n\cup\bigcup_{\xi\in L_n}G_\xi\;.$$
Then $W=\bigcup_{n\in\omega}W_n$ is open in $X$, and $W=\pi^{-1}\big[\pi[W]\big]$, so $\pi[W]$ is an open nbhd of $x'$ in $X'$. Let $H_0=\operatorname{cl}_XW_0$. Given $H_n$ for some $n\in\omega$, let $M_n=\{\xi<\kappa:q_\xi\in\pi[H_n]\}$, and let
$$H_{n+1}=H_n\cup\bigcup_{\xi\in M_n}\operatorname{cl}_XG_\xi\;.$$
Then $H=\bigcup_{n\in\omega}H_n$ is closed in $X$, and $\pi^{-1}\big[\pi[H]\big]=H$, so $\pi[H]$ is closed in $X'$. Clearly $\pi[W]\subseteq\pi[H]\subseteq U$, so
$$x'\in\pi[W]\subseteq\operatorname{cl}_{X'}\pi[W]\subseteq\pi[H]\subseteq U\;.$$
I leave it to you to modify the argument slightly to show that $X'$ is $T_3$ at the point $p^*$ as well.