Show that there is either an increasing sequence or a decreasing sequence of points $x_n$ in A with $lim_{n\rightarrow \infty} x_n=a$.

Because $a\in A \subset \mathbb{R}$ is an accumulation point, for every $n \in \mathbb{N}$ there is an element $x_n \in A$ such that $0<|x_n-a|<\frac{1}{n}$.

Then, either infinite elements of this sequence are less than $a$, or infinite elements are larger than $a$, or both.

In the first case, we build an increasing subsequence by choosing all $x_{n_k}$ elements being less than $a$ and such that $x_{n_k}<x_{n_{k+1}}$. We can do this by choosing $x_{n_{k+1}} \in (x_{n_k},a)$, which can be done by taking $n_{k+1}$ large enough so that $\frac{1}{n_{k+1}} < |x_{n_k}-a|$.

In the second case, we build a decreasing subsequence by choosing all $x_{n_k}$ elements being larger than $a$ and such that $x_{n_k} > x_{n_{k+1}}$. We do this again by choosing $x_{n_{k+1}} \in (a,x_{n_k})$, which can be done again by taking $n_{k+1}$ large enough so that $\frac{1}{n_{k+1}} < |x_{n_k}-a|$.

And in the third case, we can do any of the two previous constructions.

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Edit 1:

In the first case, let $x_{n_1}$ be the first element of $x_n$ less than $a$. This means $n_1$ is the first index for which $x_n < a$.

Let $x_{n_2}$ be the first element of $x_n$ in the open interval $(x_{n_1},a)$. Again, this means $n_2$ is the first index for which $x_n \in (x_{n_1},a)$. This index has to exist since the sequence $x_n$ gets arbitrarily close to $a$ and there are infinite elements less than $a$.

We define the subsequence recursively: $x_{n_{k+1}}$ is the first element of $x_n$ in the open interval $(x_{n_k},a)$. Such element exists because the sequence $x_n$ gets arbitrarily close to $a$ and there are infinite elements less than $a$.