Conditions for partition of unity 
Why must a space be Hausdorff and have a countable basis in order to have a partition of unity?

My only ideas so far are that, intuitively, a Hausdorff space means that the limits of convergent sequences are unique, and for a partition of unity to exist, a family (=indexed set) of open sets cover the space, so there must be a countable basis because the set is indexed with natural numbers. 
 A: Neither may be true.

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*For Hausdorffness, consider $X = \{ 0,1 \}$ and give it either the indiscrete (trivial) topology (so that it is not T0), or the topology of the Sierpiński space (so that it is T0 but not T1).  It is easy to show that any open cover of $X$ has a partition of unity subordinate to it (the constant function $x \mapsto 1$ will do, since any open cover must contain $X$).
What is true is that if $X$ is a T1 space such that every open cover has a partition of unity subordinate to it, then $X$ is Hausdorff.  For this, consider distinct $x,y \in X$, and the open cover $\{ X \setminus \{ x \} , X \setminus \{ y \} \}$ of $X$.  Then there is a partition of unity $\mathcal{F}$ subordinate to this cover.  Pick $f \in \mathcal{F}$ such that $f (x) = a > 0$ (and $f (y) = 0$).  Then $f^{-1} [\;[0,\frac{a}{2} )\;]$ and $f^{-1} [\;( \frac a2 , 1 ]\;]$ are disjoint neighbourhoods of $y$, $x$, respectively.


*For second-countability, as mentioned in my comment above, every discrete space has the property that any open cover has a partition of unity subordinate to it (take the characteristic functions of the points), but no uncountable discrete space is second-countable.
