Why are norms continuous? [closed]

Let $(X,\left\|\cdot\right\|)$ be a normed space. We need to prove that: $$\forall (x_n):\mathbb{N}\to X\ x_n\to x\implies \left\|x_n\right\|\to \left\|x\right\|$$ Let $\varepsilon>0$ and $(x_n)$ be an arbitrary sequence in $X$ that converges to $x\in X$. Then, $$\exists N\in \mathbb{N}:n\ge N\implies \left\|x_n-x\right\|<\varepsilon$$ But $$ \left| \left\|x_n\right\|- \left\|x\right\|\right|\le \left\|x_n-x\right\|$$ by the triangle inequality. Thus, $$\exists N\in \mathbb{N}:n\ge N\implies \left| \left\|x_n\right\|- \left\|x\right\|\right|<\varepsilon$$ and we are done!

A function $f$ from a metric space to a metric space is continuous if for all $x$ in the domain, for all $\varepsilon>0$, the exists $\delta>0$ such that for all points $y$ in the domain, if the distance from $x$ to $y$ is less than $\delta$, then the distance from $f(x)$ to $f(y)$ is less than $\varepsilon$.

If $f$ is a norm, then it maps a vector space into $\mathbb R$, and the distance from $x$ to $y$ is $f(x-y)$.

In this case it suffices to take $\delta=\varepsilon$, for the following reason. Suppose the distance from $x$ to $y$ is less than $\delta=\varepsilon$. Then $f(x-y)=f(y-x)<\varepsilon$ (where the equality follows from the definition of "norm"). Now recall that norms satisfy a triangle inequality: $$ f(x) \le f(y) + f(x-y) $$ $$ f(y) \le f(x) + f(y-x) $$ So $$ f(y)-f(x) \le f(y-x)<\varepsilon\text{ and }f(x)-f(y) \le f(x-y)<\varepsilon, $$ so $$ |f(x)-f(y)|<\varepsilon, $$ i.e. $$ \Big(\text{distance from $f(x)$ to $f(y)$}\Big) <\varepsilon. $$

To keep it short and straight to the point: the norm of the normed space $(X,\|\cdot\|)$ is a continuous function because the topology you (usually) consider on $X$ is the smallest topology in which $\|\cdot\|$ is continuous. So it is continuous because we want it to be continuous.