real analysis - Show that the following function is continuous

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function which takes a convergent sequence and gives us a convergent sequence.

Show that $f$ is continuous.

So I saw a proof for this but I don't get it.

proof

We show that for all convergent sequences $a_n$ with limit $a$ it holds that $\lim_{ n \rightarrow \infty } f(a_n) = f(a)$. Let $a_n$ be convergent with limit $a$. define $b_n :=a_{\frac{n}{2}}$ if $n$ even and $b_n := a$ if $n$ odd.

Now the unclear part

The following first equation is unclear:

And the last part with "it follows that" is unclear:

$\lim_{ n \rightarrow \infty } f(b_n) = \lim_{ n \rightarrow \infty } f(b_{2n+1} ) = \lim_{ n \rightarrow \infty } f(a) = f(a)$. It follow that $\lim_{ n \rightarrow \infty } f(a_n) = a.$

Answer

The sequence $\bigl(f(b_{2n+1})\bigr)_{n\in\mathbb N}$ is a subsequence of the sequence $\bigl(f(b_n)\bigr)_{n\in\mathbb N}$ and therefore, since the limit $\lim_{n\in\mathbb N}f(b_n)$ exists, you have $$\lim_{n\in\mathbb N}f(b_n)=\lim_{n\in\mathbb N}f(b_{2n+1}).$$ So, the limit $\lim_{n\in\mathbb N}f(b_{2n})$ also exists, but $b_{2n}=a_n$ and therefore $\lim_{n\to\infty}f(a_n)$ exists (and it is equal to $f(a)$).