A function taking values in a countable product of metric spaces is uniformly continuous iff its coordinate functions are. Let $\lbrace (M_n,d_n) \rbrace _{n \in I}$ be a countable family of metric spaces and define a metric in the product $\prod _{n \in I}M_n$ as follows:
$$d(\vec{x},\vec{y})=\sum _{n \in I} \min \lbrace \frac{1}{2^n},d_n(x_n,y_n)\rbrace$$
Now, let $(X,d_X)$ be a metric space and take a function $f:X \to \prod _{n \in I}M_n$,
I'm trying to prove that if all the coordinate functions $f_n:X \to M_n$ are uniformly continuous so it is $f$ (the converse follows from the facts that the projections are u.c. and a composition of u.c. functions is also u.c.). This is what I thought:
If I want $d(f(x),f(y))<\epsilon$ it would be enought to have $d_n(f_n(x),f_n(y))<\frac{\alpha}{2^n}$ for all $n \in I$ and some fixed $\alpha <\epsilon$ because if $\alpha \geq 1$ it means that $\epsilon>1$ and in that case we are done because $d(f(x),f(y))\leq 1$ for all $x,y \in X$ and if $\alpha <1$ then we get  $d(f(x),f(y))<\alpha$.
Now, since every $f_n$ is u.c. we know there are some $\delta _n$ such that if $d_X(x,y)<\delta _n$ then $d_n(f_n(x),f_n(y))<\frac{\alpha}{2^n}$, so if I could take $d_X(x,y)<\inf_{n \in I} \delta_n$ everything would fit quite well, the only problem is that I don't see any reason to assume $\inf_{n \in I} \delta_n \not=0$, is there any reason to make that assumption? Or perhaps this is not the right approach, in that case any advice on how to solve it is welcome.
Thanks in advance

HINT: You’re not taking full advantage of the definition of $d$. Let $\epsilon>0$. Then there is an $m\in\Bbb N$ such that 
$$\sum_{n\ge m}\frac1{2^n}<\frac{\epsilon}2\;.$$
Then for any $\vec x,\vec y$ we have
$$d(\vec x,\vec y)<\frac{\epsilon}2+\sum_{n<m}d_n(x_n,y_n)\;,$$
and we’re essentially dealing with only finitely many of the metrics $d_n$, those for which $n<m$. Can you finish it off from here?
