Proof

By hypothesis, $(c_n)_{n\in\mathbb N}$ is a complex sequence. We set $c_n=a_n+i\cdot b_n$ for some real numbers $a_n,b_n$.

Assume, $(c_n)_{n\in\mathbb N}$ is a convergent complex sequence with $\lim_{n\to\infty}c_n=c$.
Then, for each $\epsilon > 0$, there exists an $N\in\mathbb N$ with \[ | c_n-c | < \epsilon\quad\quad \text{ for all }n\ge N.\]
Because for every complex number $x\in\mathbb C$, we have $|\Re(x)|\le |x|$ and $|\Im(x)|\le |x|$ by the definition complex absolute value, it follows for all $n\ge N$: \[\begin{array}{c} | a_n-a |=|\Re(c_n -c)|\le | c_n-c | < \epsilon\\ | b_n-b |=|\Im(c_n -c)|\le | c_n-c | < \epsilon\\ \end{array}\]
Thus, both real sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are convergent real sequences with $\lim_{n\to\infty}a_n = \Re( c)=a$ and $\lim_{n\to\infty}b_n = \Im ( c)=b$.

Assume, the real sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are convergent real sequences with $\lim_{n\to\infty}a_n =a$ and $\lim_{n\to\infty}b_n=b$.
Then, for each $\epsilon > 0$ there exist indices $N_a,N_b\in\mathbb N$ with \[ | a_n-a | < \frac\epsilon2\quad\quad \text{ for all }n\ge N_a\] and \[ | b_n-b | < \frac\epsilon2\quad\quad \text{ for all }n\ge N_b.\]
Set $c:=a+ib$ and set $N:=\max(N_a,N_b)$.
Then, for all $n\ge N$, it follows from the triangle inequality for the distance in the metric space of complex numbers that $$\begin{align} | c_n-c | &= |a_n+ib_n-a-ib| \nonumber\\ &= |(a_n-a)+i(b_n-b)| \nonumber\\ &\le |a_n-a|+|b_n-b|\nonumber\\ & < \frac\epsilon2+\frac\epsilon2\nonumber\\ &=\epsilon.\nonumber\end{align}$$
Therefore, $(c_n)_{n\in\mathbb N}$ is a convergent complex sequence with $\lim_{n\to\infty}c_n=c$.
∎

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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983