Is this proof okay?

Let $V,W$ be two finite normed vector spaces, $D \subseteq \mathbb V$, $f: D \to \mathbb W$ be a function and $x_0 \in D$.

> $f$ is continuous in $x_0$ $\iff$ every component $f_j$ is continuous in $x_0$.

"$\implies$":

Suppose $f$ is continuous in $x_0$ and let $\varepsilon \gt 0$. Then there exists $\delta \gt 0$, such that for all $x \in D$ with $\Vert x-x_0 \Vert \lt \delta$ we have $\Vert f(x)-f(x_0) \Vert \lt \varepsilon$. Then it follows that

$$\Vert f_j(x)-f_j(x_0) \Vert = | \pi_j(f(x))-\pi_j(f(x_0)) | = | \pi_j(f(x)-f(x_0)) | \le \Vert f(x)-f(x_0) \Vert \lt …

"$\impliedby$":

Suppose $f_j$ is continuous in $x_0$ and let $\varepsilon \gt 0$. Then there exists $\delta_j \gt 0$, such that for all $x \in D$ with $\Vert x-x_0 \Vert \lt \delta_j$ we have $\Vert f_j(x)-f_j(x_0) \Vert \lt \frac{\varepsilon}{n}$ for $j = \{1,\ldots,n\}$. Choose $\delta := \min_{1 \le j \le n} \delta_j$. Then for all $\Vert x-x_0 \Vert \lt \delta$ it follows

$$\begin{align} \Vert f(x)-f(x_0) \Vert &= \Vert \big( \pi_1(f(x)), \ldots, \pi_n(f(x)) \big)^t - \big( \pi_1(f(x_0)), \ldots, \pi_n(f(x_0)) \big)^t \Vert \\ &= \Vert \big( \pi_1(f(x)-f(x_0)), \ldots, \pi_n(f(x)-f(x_…