Solutions

For the forward direction $(\implies)$, suppose that

$$z_{n} = (x_1, y_1,x_2,y_2,x_3,y_3,\dots)$$

for any two sequences $x_{n}$ and $y_{n}$. Suppose further that $z_{n}$ converges to $l$. By the definition of convergence, we then have

$$\begin{aligned} \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N &\implies z_{n} \in V_{\epsilon}(l). \end{aligned}$$

If $N$ is even, then we know that the terms

$$(y_{\frac{N}{2}}, x_{\frac{N}{2} + 1}, y_{\frac{N}{2} + 1}, \dots)$$

are in $V_{\epsilon}(l)$. On the other hand, if $N$ is odd, it is instead the terms

$$(x_{\frac{N + 1}{2}}, y_{\frac{N + 1}{2}}, x_{\frac{N + 1}{2} + 1}, y_{\frac{N + 1}{2} + 1}, \dots)$$

that are in that neighborhood. Therefore, if we choose $N_{xy} = \frac{N}{2} + 1$, we can then be certain that all $x_{n}$ and $y_{n}$ terms are in that neighborhood provided that $n \geq N_{xy}$.
For the backward direction $(\impliedby)$, suppose that the sequences $x_{n}$ and $y_{n}$ both converge to $l$. That is,

$$\lim x_{n} = \lim y_{n} = l.$$

Now, form the sequence $z_{n}$ such that

$$z_{n} = (x_1, y_1,x_2,y_2,x_3,y_3,\dots).$$

Choose any $\epsilon > 0$; by definition, since $x_{n}$ and $y_{n}$ converge,

$$\exists N_1 \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N_1 \implies \abs{x_{n} - l} < \epsilon$$

and

$$\exists N_2 \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N_2 \implies \abs{y_{n} - l} < \epsilon.$$

If we choose $N = \max \{N_1, N_2\}$, then both strict inequalities are satisfied when $n \geq N$. Consider $N_{z} = 1 + 2(N - 1)$. This value gives us the index of the term $x_{N}$ in the sequence $z_{n}$. Therefore, $z_{n}$ is in the neighborhood for every $n \geq N_{z}$.