$(X, ||• ||) $ be a $d$ dimensional normed linear space over $K$.
$\beta = \{e_1,e_2,e_3,...,e_d\} \text {be a basis of } X$
Given any, $x\in X$ has a unique representation of the form
$x= x_1 e_1 + x_2 e_2 +... +x_n e_n (x_j \in K,\forall j\in \mathbb{N}_d ) $
Then, $(x_1, x_2, x_3,..., x_d) $ is defined to be the coordinate of $x$ with respect to $\beta$.
Question : $$\text{ Given any sequence} (x^{(n)}) _{n\in \mathbb{N}}\text{ and } x \text{ in } X $$
$(x^{(n)})$$ \text{ converges to } x \text { iff it converges Co-ordinatewise. i.e } (x_j) ^{n}\to (x_j) \space{ } \forall j \in \mathbb{N}_d$
Suppose $x^{(n)}_{j} \to x_{j}$, $j=1,...,d$. Let $x := (x_{1},...,x_{d})$ and notice that: $$x^{(n)} - x = (x_{1}^{(n)}-x_{1})e_{1}+\cdots (x_{d}^{(n)}-x_{d})e_{d}.$$ Now, we have: $$||x^{(n)}-x|| \le \sum_{j=1}^{d}||(x_{j}^{(n)}-x_{j})e_{j}|| = \sum_{j=1}^{d}|x_{j}^{(n)}-x_{j}|$$ where I'm assuming the basis is normalized, so the implication follows.
Conversely, because all norms in finite dimensional spaces are equivalent, we can use the Euclidian norm: $$||x||^{2} := |x_{1}|^{2}+\cdots +|x_{d}|^{2}.$$ Suppose $x^{(n)} \to x := (x_{1},...,x_{d})$. Then: $$|x_{j}^{(n)}-x_{j}|^{2} \le \sum_{j=1}^{d}|x_{j}^{(n)}-x_{j}|^{2} = ||x^{(n)}-x||^{2}$$