consider $\displaystyle \sum_{n=1}^\infty (-1)^{n-1}a_n$ where $(a_n)$ is a monotone decreasing sequence of nonnegative numbers with $a_n \rightarrow 0$ by the alternating series test, series of this form always converge.

Show that $0 \leq \displaystyle\sum_{n=1}^\infty (-1)^{n-1}a_n \leq a_1$

There is a hint in the question:

if $(S_N)$ denotes the sequence of partial sums, consider the subsequences $(S_{2N})$ and $(S_{2N-1})$ can you show that one is decreasing while the other is increasing?

I have proven the hint - however I am unable to proceed from here - Could someone please direct me to the correct direction?

We will show that $0\leq S_{n}\leq a_{1}$ for all $n\in\mathbb{N}$ (and hence $0\leq\lim S_{n}\leq a_{1}$).
We know that $0\leq a_{1}-a_{2}=S_{2}\leq S_{1}= a_{1}$ and you already showed that $S_{2n}$ is increasing while $S_{2n-1}$ is decreasing.
Now let $k\in\mathbb{N}$ and notice that $S_{2k}\leq S_{2k-1}$. Because $S_{2n}$ is increasing and $S_{2n-1}$ is decreasing we now know that $0\leq S_{2}\leq S_{2k}\leq S_{2k-1}\leq S_{1}=a_{1}$. To finish note that, for any $n\in\mathbb{N}$, $S_{n}=S_{2k}$ or $S_{n}=S_{2k-1}$ for some $k\in\mathbb{N}$.
$$S_{n}=a_1\underbrace{-a_2+a_3}_{\leq 0}\underbrace{-a_4+a_5}_{\leq 0}\cdot\cdot \cdot\underbrace{-a_{n-1}+a_{n}}_{\leq 0}\leq a_1$$.
On the other hand if n is even than $$S_{n}=a_1\underbrace{-a_2+a_3}_{\leq 0}\underbrace{-a_4+a_5}_{\leq 0}\cdot\cdot \cdot\underbrace{-a_{n-2}+a_{n-1}}_{\leq 0}\underbrace{-a_{n}}_{\leq 0}\leq a_1$$ So in any case $S_n\leq a_1$