Is $f$ sequentially continuous? Let $f\colon \mathbb{R}\to\mathbb{R}$ be a function. Aussume that for all sequences $(x_n)_{n\in\mathbb{N}}\subset\mathbb{R}$ with $x_n\to x$ in $\mathbb{R}$ there exists a subsequence $(x_{n_k})_k$ with $f(x_{n_k})\to f(x)$ for $k\to \infty$. Does it follow that $f$ is sequentially continuous? 
Under these assumptions, one has to check that $f(x_n)\to f(x)$ for $n\to \infty$. However, I don't know how to do it. Any help will be appreciated.  
 A: Yes.
Assume that $x_n\to x$ while $f(x_n)\to f(x)$ is not true. 
Then some $\epsilon>0$ exists together with a subsequence $(x_{n_k})$ such that $|f(x_{n_k})-f(x)|\geq\epsilon$ for every $k$.  
This sequence $(x_{n_k})_k$  satisfies $x_{n_k}\to x$ but cannot have a subsequence $(x_{n_{k_i}})$ with $f(x_{n_{k_i}})\to f(x)$.
So a contradiction is found.
A: As a general rule if you have a sequence $(y_n)$ such that for any subsequence $(y_{n_k})$ there exists a further subsequence $(y_{n_{k_j}})$ that converges to $y_0$ then the original sequence, $(y_n)$, converges.
Proof. It should be clear that if $(y_n)$ is to converge, it must converge to $y_0$. The criterion for convergence says that for any $\varepsilon > 0$,
$$ y_n \in (y_0 - \varepsilon, y_0 + \varepsilon) $$
for all $n \ge$ some $N$.
If this is not the case, then there is some bad $\varepsilon_0$ such that for infinitely many $n$, 
$$ y_n \notin (y_0 - \varepsilon, y_0 + \varepsilon). $$
But now, because we have infinitely many $n$'s, we obtain a subsequence $(y_{n_k})$ with
$$ y_{n_k} \notin (y_0 - \varepsilon, y_0 + \varepsilon), \forall k. \tag{$*$} $$
On the other hand, our hypothesis is that we can take a subsequence of $y_{n_k}$ that converges to $y_0$. But this is impossible if $(*)$ holds. This finishes the proof.
This result implies what you want. Let $x_n \to x$ and put $y_n = f(x_n)$. Then if $(x_{n_k})$ is a subsequence then it is in particular a sequence and by hypothesis there is a subsubsequence $(x_{n_{k_j}})$ such that $y_{n_{k_j}} = f(x_{n_{k_j}}) \to y_0 = f(x)$. Thus by the result above, $y_n = f(x_n) \to y_0 = f(x)$.
