Is the arbitrary product of metric spaces a metric space?

No. Every metric space is first-countable, and e.g. $\mathbb{R}^\mathbb{R}$ is not first-countable.

Going to spoil the surprise: the metric Amitesh Datta defined is the categorical product in a suitable category of metric spaces, but

• the morphisms in this category need to be weak contractions: that is, if $f : M \to N$ is such a morphism, then we need $d_N(f(x), f(y)) \le d_M(x, y)$;
• the metrics in this category need to take the value $\infty$; and
• the induced topology is not the product topology.

The first requirement makes isomorphisms in this category the same as isometric isomorphisms, which is not true of the usual category of metric spaces and continuous maps (which should really be called the category of metrizable spaces). The usual category of metric spaces and continuous maps is somewhat poorly behaved, e.g. categorical constructions in this category do not come with natural metrics even when they exist.

The second requirement is very natural, e.g. it also allows colimits in a nice way. See Lawvere metric space for details and this blog post for some more details about the special case of Banach spaces (although here we don't allow infinite values of the metric and so we need to restrict our attention to a bounded version of the product).

You could define a metric space $(X,d)$ by the rule $d(x,y)=\sup_{\lambda\in \Lambda}{d_{\lambda}(x_{\lambda},y_{\lambda})}$.

Exercise 1: Is $(X,d)$ a metric space, i.e. are the axioms of a metric space satisfied? (I'm not saying whether they or or they aren't; it's an exercise to figure it out!)

It's also pertinent which topology you'd like to induce on the product. If $\Lambda$ is uncountable, then there's no guarantee that the usual product topology on $X$ is metrizable. E.g., see xyzzyx's excellent counterexample.

I hope this helps!

Edit: See Qiaochu's excellent answer below in which he discusses the $d$ I defined above in more detail!