Theorem: Scalar Multiplication on Normed Vector Space is Continuous

Let $\struct { K, +, \circ_K }$ be a normed division ring with norm $\norm {\, \cdot \,}_K$.
Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space over $K$.
Let $\struct {K \times X, \norm {\, \cdot \,}_P }$ be the direct product of $K$ and $X$ with the direct product norm $\norm {\, \cdot \,}_P$. 
Let $\circ : K \times X \to X$ be the scalar multiplication defined on $X$.
Then $\circ: K \times X \to X$ is a continuous mapping.

Proof:
Let $\tuple { \lambda_0 , x_0 } \in K \times X$.
Let $\epsilon' \in \R_{>0}$.
Set $\epsilon = \map \min { \epsilon', 1 }$
To show that $\circ$ is continuous, let $\tuple { \lambda , x } \in K \times X$ such that $\norm { \lambda_0 - \lambda }_K < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$, and $\norm { x_0 - x }_X < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$.
By definition of direct product norm, it follows that:
:$\norm { \tuple {\lambda_0, x_0} - \tuple {\lambda, x} }_P = \map \max {\norm {\lambda_0 - \lambda}_K, \norm {x_0 - x}_X } < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$.
To show that $\circ$ is continuous at $\tuple {\lambda_0, x_0}$, we calculate:
{{begin-eqn}}
{{eqn | l = \norm { \lambda_0 \circ x_0 - \lambda \circ x }_X
      | r = \norm { \lambda_0 \circ x - \lambda_0 \circ x_0 - \lambda \circ x + \lambda \circ x_0 + \lambda_0 \circ x_0 - \lambda \circ x_0 + \lambda_0 \circ x_0 - \lambda_0 \circ x}_X
      | c = adding and subtracting the same quantities
}}
{{eqn | r = \norm { \paren{ \lambda_0 - \lambda } \circ \paren  { x - x_0 } + \paren{ \lambda_0 - \lambda } \circ x_0 + \lambda_0 \circ \paren { x_0 - x } }_X
      | c = {{Vector-space-axiom|1}} and {{Vector-space-axiom|2}}
}}
{{eqn | o = \le
      | r = \norm { \paren{ \lambda_0 - \lambda } \circ \paren  { x - x_0 } }_X + \norm { \paren{ \lambda_0 - \lambda } \circ x_0 }_X + \norm { \lambda_0 \circ \paren  { x_0 - x } }_X
      | c = {{NormAxiomVector|3}}
}}
{{eqn | r = \norm { \lambda_0 - \lambda }_K \norm { x - x_0 }_X + \norm { \lambda_0 - \lambda }_K \norm { x_0 }_X + \norm { \lambda_0 }_K \norm { x_0 - x }_X
      | c = {{NormAxiomVector|2}}
}}
{{eqn | o = <
      | r = \paren { \dfrac { \epsilon }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X } }^2+ \dfrac{ \epsilon \norm { x_0 }_X }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X } + \dfrac { \epsilon \norm { \lambda_0 }_K }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }
      | c = by choice of $\lambda$ and $x$
}}
{{eqn | o = \leq
      | r = \dfrac { \epsilon }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X } + \dfrac{ \epsilon \norm { x_0 }_X }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X } + \dfrac { \epsilon \norm { \lambda_0 }_K }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }
      | c = as $\epsilon \le 1$. and $1 + \norm { \lambda_0 }_K + \norm { x_0 }_X \ge 1$
}}
{{eqn | r = \dfrac { \epsilon \paren{1 + \norm { \lambda_0 }_K + \norm { x_0 }_X} }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }
}}
{{eqn | r = \epsilon
}}
{{eqn | o = \le
      | r = \epsilon'
}}
{{end-eqn}}
It follows that $\circ$ is continuous from $K \times X$ to $X$.
{{qed}}