How to prove the sequence $a_0, a_1, \ldots$ converges iff $a_0, a, a_1, a \ldots$ converges? Problem

Prove that the sequence $a_0, a_1, a_2, \ldots$ converges to $a$ if and only if the 
  sequence $a_0, a, a_1, a, a_2, a, a_3, \ldots$ converges.

Here is my approach:
$\Rightarrow$:
            Since $a_0, a_1, a_2, \ldots$ converges to $a$, by definition of limit,
            for every $\epsilon > 0$, $\exists N \in \mathbb{N}$ such that for all 
            $n > N$, then $|a_n - a| < \epsilon$. Now consider the subsequence
            $$a, a, a, a, \ldots$$
            We have that $|a - a| < \epsilon, \, \, \forall \epsilon > 0$, thus
            $a, a, a, a \ldots$ also converges to $a$. Hence,
            $a_0, a, a_1, a, a_2, a, a_3, \ldots$ converges.
$\Leftarrow$:
            Suppose that $a_0, a, a_1, a, a_2, a, a_3, \ldots$ converges to $L$, 
            $L \neq \pm \infty$, by definition of limit, for every $\epsilon > 0$, $\exists N \in \mathbb{N}$ 
            such that for all $n > N$, then $|a_n - a| < \epsilon$, thus there must be a sequence
            $$a_{N+1}, a, a_{N+2}, a, a_{N+3}, a, a_{N+4}, \ldots$$
            that is getting closer and closer to $L$. But there is always an alternating 
            $a$ between each $a_i$ and $a_{i+1}$, so $L = a$ otherwise $|a_n - L| < \epsilon$ would
            make no sense. Therefore $a_0, a_1, a_2, \ldots$ converges to $a$. 
However I still feel it's not complete because all my reasons were based on the definition of infinite sequence. I think there must be a way to give a strong argument for this problem. I wonder if anyone could give me a hint/suggestion on my solution? Thanks.
 A: Suppose for every $\epsilon>0$, there is a  positive integer $N$ such that for every $n>N, |a_n-a|<\epsilon$. Let $(b_n)$ be the sequence $(a_0,a,a_1,a,\ldots)$. Then for every $n>2N+1$, it is clear that $|b_n-a|<\epsilon$.
Conversely, if $(b_n)$ converges, then all subsequences must converge to the same limit. Since $(a,a,a,\ldots)$ converges to $a$, the sequence $(a_n)$ does as well.
A: Your $\Rightarrow$ proof is erroneous, or at least confused. You can't show that a sequence converges by selecting a convergent subsequence from it; if you could, all sorts of things would converge, such as $\langle 0,1,0,2,0,3,0,4\ldots\rangle$.
Try it like this.   We want to show that $\langle a_0, a, a_1, a, a_2, a\ldots\rangle$ converges.  Let's call this sequence $\langle b_0, b_1, b_2, \ldots\rangle$. Let $\epsilon>0$ be given. It suffices to show that we can find $N_b$ such that for every $M_b>N_b$, $|b_{M_b} - a|<\epsilon$.

  Since $a_i$ converges to $a$, we can find an analogous $N_a$ such that   $|a_{N_a} - a|<\epsilon$ for every $n_a>N_a$.  Then take $N_b = N_a$.  If $n_b > N_b$, then either $b_{n_b}$ is $a$, and so has $|a_{n_a} - a| = 0 <\epsilon$, or $b_{n_b} = a_m$ for some $m > N_b = N_a$, and for that reason we know that $|a_m - a| <\epsilon$.

Your $\Leftarrow$ proof is also worrying me, because it seems to me that you gave up and waved your hands at the point where you say "otherwise $|a_n - L| < \epsilon$ would make no sense."  That's not a proof, and you can do better.  Why will $|a_n - L| < \epsilon$ fail for any $L\ne a$? Can you exhibit a value of $\epsilon$ that makes it fail?
But even then you aren't done. You've shown that $\langle a_0, a, a_1, a, a_2, a\ldots\rangle$ converges to $a$; now you have to show that $\langle a_0, a_1, a_2\ldots\rangle$ converges to $a$. You didn't prove this; you only asserted it. To prove it, you need to give a reason. You claim that for any given $\epsilon$, sufficiently far $a_i$ will have $|a_i - a| < \epsilon$; you gave no indication of how you planned to do this.
