$\prod_{n=1}^{\infty}{\mathbb{R}}$ endowed with the box topology is not first countable.

So assume you have a sequence $(z_n)$ in $X^+$. Let $\textbf{0}$ be the all $0$ function $\Bbb N \to \Bbb R$. Each $z_n$ is also a function $\Bbb N \to \Bbb R$ (also known as a real sequence) such that $z_n(k)>0$ for each $k \in \Bbb N$ (that's because it is in $X^+$).

Now we can apply a "diagonal argument" by defining $O = \bigcap_n (-\frac12 z_n(n), \frac12 z_n(n))$ which is a well-defined box-open neighbourhood of $\textbf{0}$ (as all $z_n(n) >0$ and so each interval is open and contains $0$). Now it's clear that $z_n \notin O$ for all $O$ (why?) and so $z_n$ cannot converge to $\textbf{0}$ at all.

It's not good enough to work with the $(-\frac1n, \frac1n)$ interval: It's possible that all $z_n$ lie in your $U$ when they're smaller than that. The "problem" neighbourhood has to depend on the sequence you're considering. I just showed that whatever the sequence is, there is a neighbourhood of $\textbf{0}$ that misses all terms of the sequence, a sort of "anti-convergence".

Note that the sequence argument disproves first countability indirectly: it is true that $\textbf{0} \in \overline{X^+}$ but in a first countable space this would imply there is a sequence from $X^+$ converging to $\textbf{0}$, while we've shown the latter not to be the case. It follows that $X$ is not first countable (as a consequence of first countability fails).