The cartesian product $M\times N$ is complete if, and only if $M$ and $N$ are complete.
My approach: Let $M$ and $N$, complete metric space, then we take a cauchy sequence $z_{n}=\{x_{n},y_{n}\}\in M\times N$. And now, the projection $p_{1}:M\times N\to M$ and $p_{2}:M\times N\to N$, are uniformly continuous, since they're lipschitz functions. In particular all uniformly continuous application transform, Cauchy sequences in Cauchy sequences. Therefore, $p_{1}(z_{n})=x_{n}$ and $p_{2}(z_{n})=y_{n}$ are cauchy sequences. So, $M\times N$ is a complete metric space.
In other direction (I can't end this), if $M\times N$ is a complete metric space, then exist the limit of sequence $z_{n}=(x_{n},y_{n})\in M\times N$, say $\lim z_{n}=z$. And for the definition of convergence sequences, for all $\epsilon>0$, we can find $n_{0}\in\mathbb{N}$, such that for all $n>n_{0}\implies d(z_{n},z)<\epsilon$. (Here is my question) And if I take the metric $d(z,z')=d(x,x')+d(y,y')$ in $M\times N$, with $z=(x,y)$ and $z'=(x',y')$, we have, $d(z_{n},z)=d(x_{n},x)+d(y_{n},y)<\epsilon$, this is sufficiently to say $\{x_{n}\}\to x$ and $\{y_{n}\}\to y$??. What happens if I take the metric $d(z,z')=\sqrt{d(x,x')^{2}+d(y,y')^{2}}$?? Is the same?