$\mathbf{Question: }$ Let $f_n: \mathbb{R} \to \mathbb{R}$. If for arbitrary $\{x_n\} \to x$, $\{f_n(x_n)\} \to f(x)$, prove $f(x)$ is continuous on $\mathbb{R}$.
Attempt:
Construction: We have $\{f_n(x_i)\}_{n=1}^\infty \to f(x_i)$ [Consider the constant sequence $\{x_i\}_{n=1}^\infty \to x_i$], for any $i \in \mathbb{N}$.
Now, for definiteness, choose $\epsilon=\frac{1}{2}>0$.
1st Step: Since $\{f_n(x_1)\}_{n=1}^\infty \to f(x_1)$, we get a $w_1 \in \mathbb{N}$, such that $|f_n(x_1)-f(x_1)|<\frac{1}{2}$ $\ \forall n \geq w_1$. Choose $n=n_1$.
m th $(>1)$ Step: Choose $\epsilon=(\frac{1}{2})^m>0$. We get $w_m \in \mathbb{N}$, such that $|f_n(x_m)-f(x_m)|<(\frac{1}{2})^m$ $\ \forall n \geq \max\{w_1,w_2,...w_m\} \geq w_m$. Choose $n=n_m>n_{m-1}>...>n_1$.
We now define a sequence $\{y_t\}_{t=1}^\infty$ such that
$$y_t=\begin{cases} x \text{ for } t \neq n_k \\x_k \text{ for } t=n_k \end{cases} $$.
Evidently $\{y_t\}_{t=1}^\infty \to x$. Thereby $\{f_t(y_t)\}_{t=1}^\infty \to f(x) $. Its subsequence $\{f_{n_k}(y_{n_k})\}_{k=1}^\infty \equiv \{f_{n_k}(x_k)\}_{k=1}^\infty \to f(x) $.
To prove: For any $\{x_p\} \to x, \{f(x_p)\} \to f(x)$.
Consider an arbitrary $\varepsilon>0$. $|f(x)-f(x_p)|\leq |f(x)-f_{n_p}(x_p)|+|f_{n_p}(x_p)-f(x_p)|$.
By construction, $\forall \varepsilon>0$, we can find an $\alpha_1 \in \mathbb{N}$ such that $|f_{n_p}(x_p)-f(x_p)|<(\frac{1}{2})^p\leq(\frac{1}{2})^{\alpha_1}<\epsilon/2$, $\forall p \geq \alpha_1$.
Since $\{f_{n_p}(x_p)\}_{p=1}^\infty \to f(x) $, $\forall \varepsilon>0$, $\exists \alpha_2 \in \mathbb{N}$ such that $|f_{n_p}(x_p)-f(x)|<\varepsilon/2$ $\ \forall p\geq \alpha_2$.
Therefore, $\forall \varepsilon>0$, $\ \exists \alpha=\max\{\alpha_1,\alpha_2\}$, such that $|f(x)-f(x_p)|<\varepsilon$ , $\ \forall p\geq \alpha$.
Hence, we conclude that $f(x)$ is continuous on $\mathbb{R}$, $x$ and $\{x_n\}$ being arbitrary.
Is the procedure correct?
Kindly $\mathbf{VERIFY}$
