The complex numbers $\textbf{C}$ with the distance $d(w,z) = |w - z|$ form a metric space. If $(z_{n})_{n=0}^{\infty} = (a_{n},b_{n})_{n=0}^{\infty}$ is a sequence of complex numbers, and $z = (a,b)$ is another complex number, then we have $\displaystyle\lim_{n\rightarrow\infty}z_{n} = z$ in this metric space iff $\displaystyle\lim_{n\rightarrow\infty}a_{n} = a$ and $\displaystyle\lim_{n\rightarrow\infty}b_{n} = b$.
My solution
Let us suppose first that $z_{n}\to z$. For every $\varepsilon > 0$, there corresponds an $N\geq 0$ such that \begin{align*} n\geq N \Rightarrow |z_{n} - z| = \sqrt{(a_{n} - a)^{2} + (b_{n} - b)^{2}} < \varepsilon \end{align*} since $|a_{n} - a| \leq |z_{n} - z|$ and $|b_{n} - b| \leq |z_{n} - z|$, we conclude that $a_{n}\to a$ and $b_{n}\to b$.
Conversely, let us suppose that for every $\varepsilon > 0$ there are $N_{1}\geq 0$ and $N_{2}\geq 0$ such that \begin{align*} \begin{cases} n\geq N_{1} \Rightarrow |a_{n} - a| < \varepsilon/2\\\\ n\geq N_{2} \Rightarrow |b_{n} - b| < \varepsilon/2 \end{cases} \end{align*}
Thus if we choose $N = \max\{N_{1},N_{2}\}$, we may claim that for every $\varepsilon > 0$ there corresponds an $N\geq 0$ such that \begin{align*} n\geq N \Rightarrow |z_{n} - z| \leq |a_{n} - a| + |b_{n} - b| < \varepsilon/2 + \varepsilon/2 = \varepsilon \end{align*} and we are done.
Am I reasoning correctly? Could someone provide another way to prove it?