How will Urysohn metrization theorem not hold if there is no countable basis, i.e. all basis are uncountable From what I read, the Urysohn metrization theorem states that a regular space $X$ with a countable basis (which is a normal space since the base is countable) is metrizable. The proof uses the Urysohn lemma whereby a function $f$ can be constructed s.t. given closed subsets $A$ and $B$ of $X$, we have $f(x)=a$ if $x \in A$ and $f(x')=b$ if $x' \in B$. 
I understand the proof (which uses a set of open sets containing $A$ and which get nearer to $B$, along with their g.l.b.). 
But just to check, what happens if we drop the assumption that there is a countable basis? How will the theorem not hold?
 A: Urysohn's lemma is 

Let $X$ be a normal space. If $A$ and $B$ are disjoint non-empty closed sets then 
  there is a continuous $f:X \to [0,1]$ such that $f[A] = \{0\}$ and $f[B] = \{1\}$.

Note there is no mention of a countable base. This holds for all normal spaces, and is one of the main reasons that normal spaces are so important: they have enough continuous real-valued functions, which opens up all sort of other facts, like embedding theorems.
E.g. Urysohn's lemma applies in all metric spaces (though we can construct enough continuous functions based on $d$ already most of the time), linearly ordered spaces in the order topology (so-called LOTS), all compact Hausdorff spaces, to name the most common classes.  All regardless of second countability.
There is a separate fact, discussed here, that a regular second countable space is normal (and this then allows Urysohn's lemma to apply). This is a totally seperate fact. In fact, a regular Lindelöf space already is normal and Lindelöf is much weaker than second countable, so that fact is even more useful. 
So your assumption that the theorem must fail in the absence of a countable base is unfounded. Urysohn's lemma is about normality.
