If I understand the question, then you are correct, there is such a space. I'll sketch what I hope is a correct argument.

Take $X=\coprod_AY$ for some fixed locally compact Hausdorff space $Y$ and some index set $A$. As long as $Y$ is sufficiently complicated (probably $Y=\mathbb R$ would work) and $A$ is sufficiently large ($\left|A\right|\geq\aleph_1$ is enough), then you can find a collection of Borel sets $\{B_\alpha\subseteq Y\}_{\alpha\in A}$ which are not all contained in any countable "stage" towards the Borel $\sigma$-algebra of $Y$ (I'm sorry I don't know the standard terminology for this; using the notation of the wikipedia entry on Borel sets, I mean that for any countable ordinal $m$, not all of the $B_\alpha$ are contained in $G^m$). Now the subset:
$$\coprod_{\alpha\in A}B_\alpha\subseteq\coprod_{\alpha\in A}Y$$
is not a Borel set. If it were, it would be contained in some $G^m(X)$ for some countable ordinal $m$, but this would imply that every $B_\alpha$ is in $G^m(Y)$, a contradiction.

If we take $Y$ metrizable, then $X$ will be metrizable as well, and hence perfectly normal.