Sequentially continuous for monotone sequences implies sequentially continuous. Is it true that if  $f:X\rightarrow Y$ satisfies that $x_n\rightarrow x\implies f(x_n)\rightarrow f(x)$ for every monotone sequence $(x_n)_n$ then $f$ satisfies this for every sequence?
(Here monotone means that $d(x_n,x)$ is a decreasing function of $n$.)
Even $X=Y=\mathbb{R}$ would be useful.
Thank you!
 A: Here's an argument for $X=\mathbb{R}$.
Suppose $(x_n)$ is a sequence in $\mathbb{R}$ that converges to $x$. We argue $f(x_n)\to f(x)$ by proving that every subsequence of $(f(x_n))$ has a further subsequence which converges to $f(x)$.
To this end, fix a subsequence $(f(x_{n_k}))$ of $(f(x_n))$. Take a monotone subsequence $(x_{n_{k_j}})$ of $(x_{n_k})$ (see: here). Then $x_{n_{k_j}}\to x$, so by hypothesis, $f(x_{n_{k_j}})\to f(x)$.
Since every subsequence of $(f(x_n))$ has a further subsequence converging to $f(x)$, we conclude $f(x_n)\to f(x)$.
A: Given $x_n \to x$, not necessarily monotone. We are to show that $f(x_n) \to f(x)$.
We use a fact here:

For a sequence $a_n$ and a point $a$, $a_n \to a$ if and only if for every subsequence $a_{n_k}$ of $a_n$, there is a subsequence $a_{n_{k_l}}$ of $a_{n_k}$ which converges to $a$.

We will try and use this on $f(x_n)$. For this, we make a claim that we shall prove:

Every sequence $x_n \to x$ has a monotone subsequence i.e. there is $x_{n_k}$ subsequence of $x_n$ which converges to $x$ monotonically.

The proof of this is easy : let $n_1$ be the smallest index $n$ for which $||x_n 
 -x|| < 1$. Then, let $n_k$ be the smallest index $n$ greater than $n_{k-1}$ for which $||x_n - x|| < \min\{||x_{n_{k-1}} - x||, \frac 1k\}$. It's clear that $x_{n_k} \to x$ monotonically.
EDIT : I don't know for which metric spaces the following above two facts are true, but I am sure they must be true over all metric spaces. I will look for a generalization.
Now, consider the sequence $f(x_n)$, and consider any subsequence $f(x_{n_k})$. Note that $x_{n_k} \to x$, so there is a monotonic subsequence $x_{n_{k_l}} \to x$ of $x_{n_k}$. However, for monotonic subsequences, it is true that $f(x_{n_{k_l}}) \to f(x)$. 
However, $f(x_{n_{k_l}})$ is a subsequence of $f(x_{n_k})$ which converges to $f(x)$. Hence, for any subsequence $f(x_{n_k})$ of $f(x_n)$, we found a convergent subsequence to $f(x)$. By the fact, it follows that $f(x_n) \to f(x)$ for any arbitrary $x_n \to x$.
Hence, the condition for monotone sequences is actually sufficient for continuity to follow. Now that I and you see it, it seems like a very strong statement.
