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The claim is false it would imply that normal spaces are countably paracompact and hence that normality of $X$ would imply normality of $X\times[0,1]$. The latter is not the case, see Mary Ellen Rudin …

Modify the proof in the paper: if $f(b)<p$ then, by continuity there is an $n$ such that $f(x,0)<p$ for all $x<-n$. The same argument as in the paper now works to establish that $f(a)\le p$. Likewise …

You can copy any standard proof of Urysohn's Lemma and substitute "member of $\mathcal{B}$" for "open set" and "complement of member of $\mathcal{B}$" for closed set. Let $\mathcal{C}$ denote $\{X\set …

Fleshing out Will Brian's suggestion (and saving him the trouble of writing it up): If $\kappa$ is finite you topologize $\lambda$ using the base $[\kappa-1,\lambda)$ (which is a perverted way of lis …

For a general source of counterexamples: look at connected but not locally connected spaces. The retracts are connected but the neighbourhoods of some points are not. The Topologist's sine curve, Knas …

For an explicit pair of disjoint open sets with intersecting closures work on the binary tree $2^{<\omega}$ of finite sequences of $0$s and $1$s. For every $x\in2^\omega$ let $A_x=\{x\mathbin{\upharpo …

To confirm Andreas' suspicion: Balcar and Vojtáš proved in Almost Disjoint Refinement of families of subsets of $\mathbb{N}$ that every ultrafilter on $\mathbb{N}$ has an almost disjoint refinement. T …