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The proof of Urysohn's lemma involves non-trivial constructions. The lemma is

If $X$ is a normal topological space, and $A,B$ are disjoint closed subsets, then there is a continuous function from $X$ to $[0,1]$ such that $f(A)=0$ and $f(B)=1$.

Edit: Assume that the normal space $X$ has property that singleton sets are closed.

(1) If $A=\{a\}$ and $B=\{b\}$ are singleton sets, they are certainly closed (as $X$ is normal, am I right?), then is it is easy to prove Urysohn's lemma? [It was not easily clicking to me, how to produce (or show existence of) a continuous function, which takes distinct values at $a$ and $b$.]

(2) If only $A=\{a\}$ is singleton, and $B$ arbitrary closed subset disjoint from $A$, still is it easy to prove?

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    For (1), it depends on your definition of normal. Some books include the $T_1$-separation axiom in their definition of normal, others don't. For a counterexample, look at any indiscrete space $X$ containing at least two elements. Then, every pair of disjoint closed subsets of $X$ will have a pair of disjoint open neighbourhoods, but singleton sets are not closed. Similarly for (2), it depends on whether you include the extra separation axiom. Just use the normal proof for the Urysohn lemma, I don't find it particularly hard, but maybe you disagree. – 2017-02-23
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    I am coming back in point-set topology after long time after my m.sc., so I have no clear ideas whether singleton sets are always closed in normal space; I looked into Wiki, where there is sequence of category of topological space $T_1, T_2, T_{2\frac{1}{2}}$, etc, and I thought $T_2$ implies $T_1$, $T_3$ implies $T_2$ and so on; thus I thought, in normal spaces, as they are $T_1$ according to this thought, singleton sets would be claosed. The other point - insted of following directly argument in standard proof, my intention was to consider first simplest cases, and think where difficulty. – 2017-02-23
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    If you take the definition of a normal space $X$ as: For all disjoint closed subsets $A$ and $B$ of $X$, there exist disjoint open subsets $U$ and $V$ of $X$ such that $A\subset U$ and $B\subset V$. Then, one point sets need not be closed as stated above. Another thing is that with all definitions I have seen it follows that $T_2\Rightarrow T_1\Rightarrow T_0$, but there is no real convention on the definition of $T_3, T_{3\frac{1}{2}}$ or $T_4$. Some of them require one point sets to be closed, others don't. – 2017-02-23

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