Proof that every normed vector space is a topological vector space

The first point is fine. For the second, fix $(v_0,\alpha_0)\in V\times K$ and $\varepsilon >0$. We have to find $\delta>0$ such that if $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$ then $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$. We have \begin{align} \lVert \alpha_0v_0-\alpha v\rVert&\leq \lVert \alpha_0v_0-\alpha v_0\rVert+ \lVert \alpha v_0-\alpha v\rVert\\ &=|\alpha_0-\alpha|\lVert v_0\rVert+|\alpha|\lVert v-v_0\rVert\\ &\leq |\alpha_0-\alpha|(\lVert v_0\rVert+\lVert v-v_0\rVert)+|\alpha_0|\lVert v-v_0\rVert. \end{align} We take $\delta$ such that $\delta^2+\delta(\lVert v_0\rVert+|\alpha_0|)\leq \varepsilon$ (which is possible).

In this case, $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$ when $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$.

Let fix $\alpha$ and $ x$ such that,

$$\|x-x_0\|<1 $$ then, $$\|x\| \le \|x-x_0\|+\|x_0\|\le \|x_0\|+1$$

\begin{align}\|\alpha x -\alpha_0x_0\| &= \|\alpha x -\alpha_0 x+\alpha_0 x-\alpha_0x_0\| \\&\le \|x\||\alpha -\alpha_0| +|\alpha_0| \|x-x_0\|\\&< (\|x_0\|+1)|\alpha -\alpha_0| +(|\alpha_0|+1) \|x-x_0\|\\&\le \max\left[(\|x_0\|+1),(|\alpha_0|+1)\right]\color{red}{\left[|\alpha -\alpha_0| + \|x-x_0\|\right]}\\&\le 2\max\left[(\|x_0\|+1),(|\alpha_0|+1)\right]\color{red}{\max\left[|\alpha -\alpha_0| , \|x-x_0\|\right]} \end{align}

for any $\varepsilon>0$ if you take $$\color{red}{\delta= \min\left(1, \frac{\varepsilon}{ 2\max(\|x_0\|+1),(|\alpha_0|+1)}\right)}$$ Then for any $x$ and $\lambda$ such that, $$\color{red}{\max\left[|\alpha -\alpha_0| , \|x-x_0\|\right] <\delta}$$ you get, $$\color{blue}{ \|\alpha x -\alpha_0x_0\|<\varepsilon}$$