Is the sum of norms/metrics a norm/metric on the product space?

Question

Let be $\mathfrak X:=\Big\{\big(X_i,\nu_i\big):i\in I\Big\}$ a FINITE collection of normed spaces. So I would like to know if the position $$ \nu(x):=\sum_{i\in I}\nu_i(x_i) $$ for all $x\in\prod_{i\in I}X_i$ defines a norm on $\prod_{i\in I}X_i$. So if $x\in \prod_{i\in I}X_i$ is such that $$ \nu(x)=0 $$ then surely $\nu_i(x_i)=0$ for all $i\in I$ because $I$ is finite and because the quantities $\nu_i(x_i)$ for $i\in I$ are not negative; moreover the distributivity law implies that $$ \nu(\lambda\cdot x)=\sum_{i\in I}\nu_i(\lambda\cdot x_i)=\sum_{i\in I}\lambda\cdot\nu_i(x_i)=\lambda\cdot\sum_{i\in I}\nu_i(x_i)=\lambda\cdot\nu(x) $$ for all $x\in X$; finally if for any $x,y\in X$ the inequality $$ \nu_i(x_i+y_i)\le\nu_i(x_i)+\nu(y_i) $$ must holds for any $i\in I$ then summing we conclude that $$ \nu(x,y)=\sum_{i\in I}\nu_i(x_i+y_i)\le\sum_{i\in I}\big(\nu_i(x_i)+\nu_i(y_i)\big)=\sum_{i\in I}\nu_i(x_i)+\sum_{i\in I}\nu_i(y_i)=\nu(x)+\nu(y) $$ So we can claim that effectively $\nu$ is a norm on $\prod_{i\in I} X_i$. So first of all I ask if effectively the funciont $\nu$ is a norm when $I$ is FINITE and so if the argumentations I gave are correct. Moreover can be the same argumentations implemented to prove an analogous statement for metric spaces? that is if $\mathfrak X:=\Big\{\big(X_i,d_i\big):i\in I\Big\}$ is a FINITE collection of metric spaces then does the position $$ d(x,y):=\sum_{i\in I}d_i(x_i,y_i) $$ for all $x,y\in \prod_{i\in I}$ define a metric on $\prod_{i\in I}X_i$? Moreover what happens if $I$ is not finite, that is what happens if $$ |I|=\aleph_\alpha $$ with $\alpha\ge 0$? I am sure that in this case generally the above positions does not define a norm or a metric but unfortunately I was not able to find a counterexample so that I thought to ask a specific question. So could someone help me, please?