real analysis - lim inf, lim sup, limit points

Let $t$ and $s$ be real numbers with $t. Suppose that as $n \to \infty$, $a_{2n} \to t$ and $a_{2n-1}\to s$.

True or False: It must be that $\liminf a_{n} = t$ and $\limsup a_{n} = s$ and $(a_n)$ has no other limit points.

Prove the statement or give a counterexample.

I think true, based on the following example, which I adapted from a wikipedia image (where s=1, t=-1 and $a_{2n-1}$ is the pink sequence and $a_{2n}$ is the light blue sequence):

Hence, all of what the statement said is true... but i don't know how to prove it.

Answer

I will use the following definition of $\limsup$ and $\liminf$:

Let $E$ be the set of subsequential limits of $(a_n)$. Then $\limsup_{n \to \infty} a_n := \sup E$ and $\liminf_{n \to \infty} a_n := \inf E$.

It suffices to show that in the setting of your question, $E = \{t, s\}$.
Clearly $t,s \in E$. Now suppose $(a_{n_k})$ is a subsequence of $(a_n)$. The sequence of indices $n_k$ must either contain infinitely many even numbers or infinitely many odd numbers (or both).

Edit: This implies that $(a_{n_k})$ contains a further subsequence that either converges to $s$ or to $t$. Thus if $(a_{n_k})$ converges, it must be to either $s$ or $t$. (But it might not converge.)

So $E = \{t,s\}$.