Proving convergence of a bizarre sequence This was a question on an exam I recently took that I didn't do so great on, so I'm trying to understand this problem thoroughly. What flaws are there?

Suppose that $\{a_n\}$ and $\{b_n\}$ are two convergent sequences that both converge to the same $L \in \mathbb{R}$. Define a new sequence $c_n$ such that $c_n = a_n$ if $n$ is odd and $c_n = b_n$ if $n$ is even. Prove that $\{c_n\}$ converges to $L$.
Proof Since $a_n$ and $b_n$ both converge to $L$, we know that
  $$\lim \sup a_n = \lim \inf a_n = \lim \sup b_n = \lim \inf b_n = L$$
  Define $i_n = \inf\{c_k : k \geq n\}$ and $s_n = \sup\{c_k : k \geq n\}$. Then $\lim i_n = \lim \inf c_n$ and $\lim s_n = \lim \sup c_n$.
$L$ is a subsequential limit of $c_n$ and so $\lim \inf c_n \leq L \leq \lim \sup c_n$.
Suppose for contradiction that $\lim \inf c_n < L$. Then there exists an $n_0$ such that $L \neq i_{n_0}$ and $i_{n_0} < a_{n_0}$ or $i_{n_0} < b_{n_0}$, but this contradicts the construction of $\{a_n\}$ and $\{b_n\}$ because one or both would not converge at all or would converge to $i_{n_0}$. Thus, $\lim \inf c_n = L$.
A similar argument can be used to show that $\lim \sup c_n = L$ by replacing $i_{n_0}$ with $s_{n_0}$, $\lim \inf c_n$ with $\lim \sup c_n$ and switching the inequality signs in the previous paragraph.
Since $\lim \inf c_n = L = \lim \sup c_n$, $c_n$ converges to $\lim a_n = \lim b_n = L$.

 A: I am not sure whether your current proof is wrong but I am confident that it has some unneeded extra parts. This proof should be $3$ or $4$ sentences.
For a simpler approach try this without invoking $\sup$ or $\inf$. If you need help see below: 

Let $\epsilon>0 $ be given we need to show that $\exists N_c$ such that $\forall n>N_c$ we have $|c_n-L|<\epsilon$ but because we know $\{a_n\}$ converges we can select $N_a$ such that $\forall n>N_a$ we have $|a_n-L|<\epsilon$ and $\{b_n\}$ converges we can select $N_b$ such that $\forall n>N_b$ we have $|b_n-L|<\epsilon$. So we can just take $N_c=\max\{N_a,N_b\}$ and then we are guaranteed to be within $\epsilon$ of our target. 

A: The other answers give quick proofs, but if you want to use sups and infs, that's ok too. But the flaw in your proof, in my opinion, starts here: "Then there exists an $n_0$....". Let's pick it up at "Suppose for contradiction that $\liminf c_n<L$", which is fine. Here is a sketch of what comes next. I will leave the details to you: if $\liminf c_n$ is strictly less than $L$, then there is an infinite number of terms of $c_n$ less than $L$. And this means that-----
A: $$c_{2n}=b_{2n}$$
$$c_{2n+1}=a_{2n+1}$$
$$\lim_{n\to+\infty}b_{2n}=\lim_{n\to+\infty}a_{2n+1}=L$$
thus
$$\lim_{n\to+\infty} c_{2n}=\lim_{n\to+\infty} c_{2n+1}=L$$
and
$$\lim_{n\to+\infty} c_n=L$$
A: HINT
As noticed, I am also not sure whether your current proof is wrong but it seems too much complicated indeed we don't need to use $\liminf$ and $\limsup$ concept since it can be carried out directly by the definition of limit.
Indeed, since $c_n \to L$ for $n$ even and $c_n \to L$ for $n$ odd, since we are considering two subsequences wich cover all the possible values for $n$, we can easily shown that $c_n \to L$.
A: Here's a simple proof
$\textbf{Proof:}$
Assume $\epsilon>0$, and by convergence choose $N_1,N_2\in\mathbb{N}$ such that $a_{n_1},b_{n_2}\in\{x\in\mathbb{C}|\text{dist}(x,L)<\epsilon\}$ for all $n_1\geq N_1$ and $n_2\geq N_2$. Then choosing $N=\max(N_1,N_2)$ we find $c_n\in\{x\in\mathbb{C}|\text{dist}(x,L)<\epsilon\}$ for all $n\geq N$, therefore $c_n\to L$, for $\epsilon$ was arbitrary.$\square$
