Let $I$ be a countable set, $E_i$ be a $\mathbb R$-Banach space for$^1$ $i\in I$, $$\prod_{i\in I}E_i:=\left\{x:I\to\bigcup_{i\in I}E_i\mid x(i)\in E_i\text{ for all }i\in I\right\}\tag1$$ and $$E:=\left\{x\in\prod_{i\in I}E_i:\sup_{i\in I}\left\|x(i)\right\|_{E_i}<\infty\right\}$$ be equipped with $$\left\|x\right\|_E:=\sup_{i\in I}\left\|x(i)\right\|_{E_i}\;\;\;\text{for }x\in E.$$
Is $E$ a (complete?) normed $\mathbb R$-vector space?
I'm a bit confused by this answer (indicating that the desired claim is not true) and section 1.3 in this lecture notes (indicating that the desired claim is true). But maybe the answer is just "yes, but the induced topology is not the product topology".
$^1$ I'm not sure if $\prod_{i\in I}E_i$ or $\times_{i\in I}E_i$ is the correct notation for the space on the right-hand side of $(1)$ (or if both are the same). Is it what is called the "direct product" or is it just the usual "cartesian product"? Would be great if someone could elaborate on that in the comments.
The answer is yes.
Note that if $\{ x^n \}^n$ is a Cauchy sequence in $E$, then, for all fixed $i$ the sequence $\{ x^n(i) \}^n$ is a Cauchy sequence in $E_i$, therefore, it converges to some $y(i)$. Let $y := y(i)$.
Next, for each $\epsilon>0$ there exists some $N$ so that, for all $n,m>N$ you have $$ \| x^n - x^m \|<\epsilon $$
Taking the limit by $m$you get $$ \| x^n - y \| \leq \epsilon $$
From here, you can easily deduce both that $y \in E$ and $x^n \to y$.
Next Note also that this norm cannot induce the product topology. The reason is simple: The unit ball is open in the norm topology. If this topology would be induced product topology, there would be some set $U \subset \prod_{i} E_i$, which is open in the product topology such that $$\{ x \in E : \|x \|_E <1 \} = E \cap U$$
Now, write what it means for $U$ to be open in the product topology to get a contradiction.