I think I have a proof, but it's a little convoluted.
Since $\tau$ is strictly finer than than $\tau_{st}$, there is an open set $U \in \tau$ such that $U \notin \tau_{st}$. For every $x \in \mathbb{R}$, the sets $U \cup (-\infty, x+1) $ and $U \cup (x, +\infty)$ are both open in $\tau$. However, at most one of the two can be in $\tau_{st}$. For the sake of contradiction, assume both are in $\tau_{st}$, then so is their union, which is $U$, which cannot be true. Thus for every $x \in \mathbb{R}$ we have constructed at least one set open in $\tau$ which is not in $\tau_{st}$, proving their difference is not countable.
Is this correct? Can it be done easier?
Best Answer
Instead of the sets you are considering look at $U\setminus \{x\}$ for $x \in \mathbb R$. You can see that $U\setminus \{x\}\neq U\setminus \{y\}$ whenever $x,y \in U$ and $x \neq y$ ; also $U\setminus \{x\} \in \tau_{st} $ for at most one $x$ (because $U =U\setminus \{x\})\cup U\setminus \{y\}))$ for $x \neq y$ and $U\setminus \{x\} \in \tau $ for all $x$. When $U$ is countable consideration of the intervals $U \cup (x,x+1)$ can be used. I leave the details to you.