For $n \in \mathbb{N} \ \ $ let $(E_1, \vert \vert * \vert \vert_{E_1}), ..., (E_n, \vert \vert * \vert \vert_{E_n}) $ be Banach spaces. Define $E:= \prod_{j=1}^{n}E_n$, $\vert \vert * \vert \vert_E : E \rightarrow \mathbb{R}, \ \ (x_1, ..., x_n) \mapsto \vert \vert (\vert \vert x_1 \vert \vert_{E_1}, ..., \vert \vert x_n \vert \vert_{E_n} )\vert \vert_p \ \ $ for $\ \ p \in [1, \infty].$
Let $\vert \vert \vert * \vert \vert \vert $ be another norm on $E$, such that for all $k \in \{1, ..., n\}$ the function $p_k: (E, \vert \vert \vert * \vert \vert \vert ) \rightarrow (E_k, \vert \vert * \vert \vert_{E_k} ), \ \ (x_1, ..., x_n) \mapsto x_k $ is continuous. Prove that the norms $\vert \vert * \vert \vert_{E}$ and $\vert \vert \vert * \vert \vert \vert$ are equivalent.
If we assume that $\lvert \lvert \lvert \cdot \rvert \rvert \rvert$ makes $E$ a Banach space then by the Open mapping theorem (or Banach's isomorphism theorem or whichever your favourite equivalent form of the OMT is), it will suffice to see that $\|x\|_E \leq c \lvert \lvert \lvert x \rvert \rvert \rvert$.
There are many ways to see this. One way is to recall that $\|\cdot\|_E$ induces the product topology on $E$ so that if $Y$ is a topological space $f:Y \to (E, \|\cdot\|_E)$ is continuous iff $p_k \circ f$ is continuous for each $k$. In particular, it is immediate that $\operatorname{Id}: (E, \lvert \lvert \lvert \cdot \rvert \rvert \rvert) \to (E, \|\cdot\|_E)$ is continuous which is precisely that $\|x\|_E \leq c \lvert \lvert \lvert x \rvert \rvert \rvert$ as desired.
Alternatively, continuity of $p_k$ means that $\|x_k\|_{E_k} \leq c_k \lvert \lvert \lvert x \rvert \rvert \rvert$. Hence there is a $c$ such that $\max_k \|x_k\| \leq c \lvert \lvert \lvert x \rvert \rvert \rvert$. To conclude exploit the fact that the $p$-norms are equivalent on $\mathbb{R}^k$.