Uniform metric topology is contained in box topology We consider the cartesian product $$X=\prod_{s\in [2,3]}[0,1]=\{(t_s)_{s\in [2,3]  }:t_s\in [0,1]\}.$$ 
I will just write $(t_s)$ instead of $(t_s)_{s\in [2,3]}$.
Let $\tau_b$ be the box topology where the open basic subsets are $\prod_{s\in [2,3]}U_s$ where each $U_s\subseteq [0,1]$ is open, and $\tau_\phi$ the topology induced by the uniform metric $$\phi((t_s),(y_s))=\sup\{|t_s-y_s|:s\in [2,3]\}.$$
I want to prove that $\tau_{\phi}\subseteq\tau_b$. I think I almost have it, but still stuck in some part.
Let $B((t_s),\epsilon)$ be an open ball in $(X,\tau_{\phi})$ and $(y_s)\in B((t_s),\epsilon)$. In order to show that $(y_s)$ is an interior point of $B((t_s),\epsilon)$ in the space $(X,\tau_b)$, I must show there is some open basic subset in the box topology between them.
First I tried with $\delta = \epsilon -\phi ((t_s),(y_s))$, and wanted to show if $$\prod_{s\in [2,3]}(y_s-\delta,y_s+\delta)\subseteq B((t_s),\epsilon).$$
But it won't work because if $(z_s)$ is in that product, then for each $s\in [2,3]$, we have $|z_s-y_s|<\delta $ and then $$|z_s-t_s|\le|z_s-y_s|+\phi((t_s),(y_s))<\epsilon,$$
and this only takes us to $\phi((z_s),(t_s))\le\epsilon$, so we cannot say that $(z_s)$ is in the open ball $B((t_s),\epsilon)$.
Any hint? Thank you.

Hint: Try using a smaller $\delta$.  If you replace $\delta$ with some $\delta'<\delta$, then at the end you'll get $|z_s-t_s|<\epsilon-(\delta-\delta')$, which is a bound on $|z_s-t_s|$ independent of $s$ which is $<\epsilon$.
