$x_n$ convergence to $x$ implies $f_n(x_n)$ convergence to $f(x)$. prove that $f$ is continuous 
Let $f$ and $f_n$ be functions from $\mathbb{R} \rightarrow \mathbb{R}$ Assume that $f_n (x_n) \rightarrow f (x)$ as $n\rightarrow \infty$ whenever $x_n \rightarrow x$. Prove that $f$ is
  continuous. (Note: the functions $f_n$ are not assumed to be
  continuous.)

Here is my approach to prove it :
Suppose $(x_n)\rightarrow x$. We want to prove $f(x_n)\rightarrow f(x)$
And perhaps we may write :
$|f(x_n)- f(x)|\leq|f(x_n)- f_n(x_n)|+|f_n(x_n)-f(x)|$
We can control the second term in RHS, but I don't think there's a way to control the first absolute value.
Another intuitive approach is working with diameter of the matrix of $f_n$ values.
 A: Suppose that $x_n\to x$ and $f(x_n)\not\to f(x)$. By passing to a subsequence, you can consider that $x_n\to x$ and $|f(x_n)-f(x)|\geq\varepsilon$ for all $n\in\mathbb N$, for some $\varepsilon>0$ fixed. Therefore, for all $m,n\in\mathbb N$, $$|f(x_n)-f_m(x_n)|+|f_m(x_n)-f(x)|\geq|f(x_n)-f(x)|\geq\varepsilon.$$ If you fix $n$, since the constant sequence $(x_n)_{m\in\mathbb N}$ converges to $x_n$, you have that $f_m(x_n)$ converges to $f(x_n)$, so there exists $m_n\in\mathbb N$ such that $|f(x_n)-f_m(x_n)|<\frac{\varepsilon}{2}$ for all $m\geq m_n$. So, for all $n\in\mathbb N$, there exists $m_n\in\mathbb N$ such that $|f_m(x_n)-f(x)|>\frac{\varepsilon}{2}$ for all $m\geq m_n$. Therefore, you can do the following: there exists a strictly increasing sequence $(m_n)_{n\in\mathbb N}$ in $\mathbb N$ such that $$\forall n\in\mathbb N,\,\,|f_{m_n}(x_n)-f(x)|>\frac{\varepsilon}{2}.$$ Construct now a sequence $y_n$ as follows: $$y_n=x_1\,\,\mathrm{for}\,\,n=1,\dots, m_1,\\ y_n=x_2\,\,\mathrm{for}\,\,n=m_1+1,\dots m_2,\\y_n=x_3\,\,\mathrm{for}\,\,n=m_2+1,\dots m_3,\,\,\mathrm{etc}.$$ Then $y_n\to x$, therefore $f_n(y_n)\to f(x)$. So, by passing to the subsequence $f_{m_n}(y_{m_n})$, you have that $$f_{m_n}(y_{m_n})\to f(x)\Rightarrow f_{m_n}(x_n)\to f(x),$$ which is a contradiction. Therefore $f$ is continuous. 
