To expand on Sourav's excellent answer. You missed a crucial point in Uryshon Metrization theorem that the space must be Hausdorff.
1.An uncountable discrete space is metrizable given by the discrete metric. It is regular as given any point $x$ and a closed set $C$. we have $\{x\}$ and $C$ are disjoint open neighbourhoods of $\{x\}$ and $C$. Similarly it is normal as given two closed sets $F,C$ they themselves are open disjoint nbds of itself. Also these are $T_{1}$.
It is not second countable as if $\mathcal{B}$ be a basis for $(X,\tau_{\text{discrete}})$ then given an open set $\{x\}$ there exist a basic open set $B$ subset of $\{x\}$ and $x\in B\implies B=\{x\}$ . Thus $\{x\}\in\mathcal{B}\,,\forall\,x\in X$. Thus $\mathcal{B}$ is uncountable.
2 .The cofinite topology on $\Bbb{N}$ is not Hausdorff as given two points $x$ and $y$. Then any open set $U$ containing $x$ is of the form $U=\Bbb{N}\setminus F$ where $F$ is some finite subset of natural numbers. Then any open set containing $y$ say $V=\Bbb{N}\setminus C$ where $C$ is some finite set must have non-empty intersection with $U$. This is because $V\cap U= \Bbb{N}\setminus(F\cup C)$ which is infinite (because $F$ and $C$ were finite). In fact you could have worked with any infinite set.
Thus it is not metrizable as it is not Hausdorff .
This set is second countable as the set of all finite subsets of a countable set is countable. Hence you can just take $\{N\setminus F\}_{F\subset\Bbb{N}\,,\text{finite}}$ to be a countable basis.
This is a Normal space as each singleton is closed so it is $T_{1}$ . And given two
disjoint closed sets $F,C$ . You have $F$,$C$ must be finite and hence $\Bbb{N}\setminus F$ and $\Bbb{N}\setminus C$ are disjoint open neighbourhoods. This also proves regularity as each singleton is clsoed.
