Yesterday, user t.b. linked me a passage in Counterexamples in Topology.
In example 75, linked above, I have a questions on properties 2 and 5.
The closure of each basis neighborhood $N_\epsilon((x,y))$ contains the union of the four strips of slope $\pm\theta$ emanating from $B_\epsilon(x+y/\theta)$ and $B_\epsilon(x-y/\theta)$.
For property 2, why is the closure of a basis neighborhood the union of those 4 strips? After that I understand why any two basis neighborhoods must intersect.
Every real-valued continuous function $f$ on $(X,\tau)$ is constant, for if $f$ were not constant, $f(X)$ would contain two disjoint open sets with disjoint closures. The inverse images would then be disjoint open sets with disjoint closures, which is impossible.
For property 5, why does the image of a nonconstant continuous function contain two disjoint open sets with disjoint closures? Since the function is continuous, I understand the inverse image of the disjoint open sets will be disjoint, but why are the closures of the inverse images also disjoint? (Essentially, is there a more explicit reason than the one Steen and Seebach give?)