Answer
Proof given below.
Work Step by Step
Let the sequence $\{c_{n}\}$ be defined as
$c_{n}=\left\{\begin{array}{ll}
a_{n}, & \text{ if n is odd}\\
b_{n} & \text{ if n is even}
\end{array}\right..$
Select any positive number $\epsilon.$
Since $\{a_{n}\}$ converges to L, then, for the chosen $\epsilon\gt 0$, there exists an index $N_{1}$ such that
for all terms of $\{a_{n}\}$ that follow ($n\gt N_{1}),\ |L-a_{n}|\lt\epsilon.$
This means that for $n\gt 2N_{1}+1$,
all odd-indexed terms $c_{n}$ are such that $|L-c_{n}|\lt\epsilon.$
Since $\{b_{n}\}$ converges to L, then, for the chosen $\epsilon\gt 0$, there exists an index $N_{2}$ such that
for all terms of $\{b_{n}\}$ that follow ($n\gt N_{2}),\ |L-b_{n}|\lt\epsilon.$
This means that for $n\gt 2N_{2}$, all even-indexed terms $c_{n}$ are such that $|L-c_{n}|\lt\epsilon.$
So, for the given $\epsilon\gt 0$, there exists an index $M=\displaystyle \max\{2N_{1}+1,2N_{2}\}$ such that
$ n\gt M\Rightarrow|L-c_{n}|\lt\epsilon$, regardless whether n is odd or even.
By definition, $\{c_{n}\}$ converges to $L.$
