$\Bbb{R}^{\Bbb{N}}_{\square}$ is not ccc

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Consider $\Bbb{Z}^{\Bbb{N}} \subseteq \Bbb{R}^{\Bbb{N}}$. The set $\Bbb{Z}^{\Bbb{N}}$ is uncountably infinite, since $|\Bbb{Z}^{\Bbb{N}}|$ = $|\Bbb{Z}|^{\Bbb{|N|}}$ = $\aleph_0^{\aleph_0}$ > $2^{\aleph_0}$ = $\frak c$.

Now let $x = (x_1, x_2,....)\in \Bbb{Z^{\Bbb{N}}}$ a sequence with all integer values and the open set containing x of the form $U = B_{\epsilon}(x_1) \times B_{\epsilon}(x_2) \times ....$, where $\epsilon < 0.5$, so that the open set U around x contains no other all integer sequence. So we have uncountably many disjoint open sets since ($\Bbb{Z}^{\Bbb{N}}$ is uncountable) and hence, the box topology is not ccc.
I'm wondering if I have the right idea and for some feedback. Thanks.