Two convergent sequences in a metric space. Question: Let {$x_n$} and {$y_n$} be two convergent sequences in a metric space (E,d). For all $n \in \mathbb{N}$, we defind $z_{2n}=x_n$ and $z_{2n+1}=y_n$. Show that {$z_n$} converges to some $l \in E$ $\longleftrightarrow$ $ \lim_{n \to \infty}x_n$= $\lim_{n \to \infty}y_n$=$l$. 
My Work: Since $x_n$ converges, $\exists N_1 \in \mathbb{N}$ s.t. $\forall n \geq N_1$, $|x_n-l_1|<\epsilon$. Likewise, since $y_n$ converges, $\exists N_2 \in \mathbb{N}$ s.t. $\forall n\geq N_2$, $|y_n-l_2|<\epsilon$. Because $z_{2n}=x_n$ and $z_{2n+1}=y_n$, then pick $N=\max\{N_1,N_2\}$. Since eventually $2n+1>2n>N$, if $z_n$ converges to $l$, then $|z_n-l|=|x_{n/2}-l|<\epsilon$ because of how we picked our N. Am I correct in this approach and should continue this way or am I wrong? This is a homework problem so please no solutions!!! Any help is appreciated. 
My work for the other way: If $\lim_{n \to \infty}x_n=\lim_{n \to \infty}y_n=l$,then we show $\lim_{n \to \infty}$$z_n=l$.Then for $N \in \mathbb{N}$, where $N=max{(\frac{N_1}{2},\frac{N_2-1}{2})}$. Since $|x_n-l|=|y_n-l|=|z_{2n}-l|=|z_{2n+1}-l|<\epsilon$ then for $n\geq N$, $|z_n-l|<\epsilon$. 

Hint  Remember the fact that "every convergent sequence is a Cauchy sequence". 
$(\Rightarrow)$ Assume $\lim_{n \to \infty} z_n=l$, then notice that
$$ |x_n-l|=|z_{2n}-l|=|(z_{2n}-z_n)+(z_n-l)|\leq |z_{2n}-z_n|+|z_n-l|<\dots $$
