Norm of Finite Product of Normed Spaces generates Product Topology Let $(X_i,||\cdot||_i)$ be normed spaces where $i=1,2,...,n \in \Bbb Z_+$.
Let $X$ be the product space of these normed spaces equipped with product topology.
For $x=(x_1,x_2,...,x_n) \in X$ where $x_i \in X_i$ $(i=1,2,...,n)$ set:
$$||x|| := \max \{||x_1||_1,||x_2||_2,...,||x_n||_n\}$$
How to prove that this norm $||\cdot||$ generates the product topology on $X$?
 A: This is true for arbitrary metric spaces of which normed spaces are just a special case.

Given metric spaces $(X_1, d_1), \ldots, (X_n, d_n)$ the product topology on $X = X_1 \times \cdots \times X_n$ is generated by the metric
  $$
d(x, y) = \max\{d_1(x_1, y_1), \ldots , d_n(x_n, y_n)\}
$$ You can verify that this is indeed a metric. Also for normed spaces $d_k(x_k, y_k)$ is just $||x_k - y_k||_k$.

For let $U = U_1 \times \cdots \times U_n$ be a basis element of the product space $X$ where each $U_k$ is open in $X_k$. And let $x = (x_1, \ldots, x_n)$ be a point in $U$. Then you can show that there is an open $d$-ball around $x$ contained inside $U$.
Indeed note that as each $U_k$ is open there are open $d_k$-balls $B_{d_k}(x_k, r_k) \subseteq U_k$ around each component $x_k$ in $X_k$. Now let $r = \min\{r_1, \ldots, r_n\}$. Then you can verify that $B_d(x, r)$ is an open $d$-ball around $x$ in $X$ such that $B_d(x, r) \subseteq U$.
Conversely let $B_d(x, r)$ be an open $d$-ball around $x$ in $X$. You can show that there is a basis element $U = U_1 \times \cdots \times U_n$ around $x$ that is contained inside that open ball.
Indeed just let $U = B_{d_1}(x_1, r) \times \cdots \times B_{d_n}(x_n, r)$ and we are done.
Hence every open set with respect to the product topology on $X$ is also an open set with respect to the above $d$-metric on $X$. And every open set with respect to the same $d$-metric on $X$ is also an open set with respect to the product topology on $X$ as expected.
A: A basic neighborhood of $0$ in the product topology has the form
$$
U=B_1(0,r_1)\times B_2(0,r_2)\times\dots\times B_n(0,r_n)
$$
where $B_i(0,r_i)=\{x\in X_i:\|x_i\|<r_i\}$. If $r=\min\{r_1,r_2,\dots,r_n\}$, then
$$
\{(x_1,\dots,x_n):\|(x_1,\dots,x_n)\|<r\}\subseteq U
$$
Conversely,
$$
B_1(0,r)\times B_2(0,r)\times\dots\times B_n(0,r)\subseteq
\{(x_1,\dots,x_n):\|(x_1,\dots,x_n)\|<r\}
$$
