Let $A \in \mathbb{R}$, $f: A \rightarrow \mathbb{R}$, and $x_0 \in A$. Then $f$ is continuous at $x_0$ $\iff$ given any monotonic sequence $\{x_n\}_{n=0}^\infty$ in $A$ with $x_n \rightarrow x_0$, we have $f(x_n) \rightarrow f(x_0)$
Here's my attempt:
"$\implies$" Suppose $f$ is continuous at $x_0$. By definition of continuity, for all sequences $\{t_n\}_{n=0}^\infty$ with $t_n \rightarrow x_0$, we have $f(t_n) \rightarrow f(x_0)$. In particular, this is true for all monotonic sequences $\{x_n\}$ in A. Hence, we're done.
"$\impliedby$" Suppose for all monotonic sequences $\{x_n\}$ in $A$ with $x_n \rightarrow x_0$, we have $f(x_n) \rightarrow f(x_0)$.
This is where I'm stuck, because I don't know how to go from monotonic sequences to all sequences. Intuitively, if we define piecewise $x_n = 1/n$ if $n$ is even and $x_n=-1/n$ if $n$ is odd, then this is a non-monotonic sequence with two monotonic subsequences $y_k = \frac{1}{2k}$ and $z_k = \frac{1}{2k+1}$ that converge to $0$, and so, by assumption, $f(y_k) \rightarrow f(0)$ and $f(z_k) \rightarrow f(0)$ so we can write up an $\epsilon - N$ proof that shows $f(x_n) \rightarrow f(0)$.
But is it generally true that all convergent sequences can be broken down to convergent monotonic subsequences? How do I make this rigorous?
Thanks!
Edit: Definition of convergence
$(a_n)$ is said to converge to $l \in \mathbb{R}$, if $\forall \epsilon >0$, $\exists N \in \mathbb{N}$ s.t. $n \geq N \implies |a_n -l| < \epsilon$.