I'm proving that on a normed space $(X, \Vert.\Vert)$, the function $\mathbb{R} \times X \to X: (\lambda, x ) \mapsto \lambda x$ is continuous on its domain, where the domain is equiped with the maximum norm $d_M((a,x),(b,y)) = \max\{|a-b|,\Vert y - x \Vert\}$
I fixed a point $(\lambda_0,x_0)$. I have to prove that:
$$\forall \epsilon > 0: \exists \delta > 0: \forall (\lambda, x) \in \mathbb{R}\times X: \max\{|\lambda- \lambda_0|, \Vert x - x_0\Vert \} < \delta \implies \Vert\lambda x - \lambda_0 x_0\Vert < \epsilon$$
I did the following:
$$\Vert \lambda x - \lambda_0 x_0\Vert \leq (|\lambda - \lambda_0| + |\lambda_0|)\Vert x - x_0\Vert + \Vert x_0\Vert |\lambda - \lambda_0| < (\delta+ |\lambda_0|)\delta + \Vert x_0 \Vert \delta$$
Can I choose $$\delta := \min\left\{\sqrt{\frac{\epsilon}{2}}, \frac{\epsilon}{2(|\lambda_0| + \Vert x_0 \Vert + 1)}\right\}$$
to make the expression smaller than $\epsilon$?
Choose $\delta < \min(1, \frac{\epsilon}{3},\frac{\epsilon/3}{|\lambda_0|}, \frac{\epsilon/3}{\|x_0\|})$. (Be careful if $\lambda_0=0$ or $\|x_0\|=0$.) Then
$$\delta^2+|\lambda_0|\delta+\|x_0\|\delta < \frac{\epsilon}{3}\cdot1+|\lambda_0|\cdot\frac{\epsilon/3}{|\lambda_0|}+\|x_0\| \cdot \frac{\epsilon/3}{\|x_0\|}= \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} = \epsilon.$$
In general, when you are finding a value of $\delta$ to make an expression less than $\epsilon$, you don't have to solve for $\delta$ exactly. What I did above was to make $\delta$ less than several different quantities, because I wanted to use different bounds in different places.