What is the right way to show that $\lim\limits_{n \to \infty} f_n(x_n) = f(x), x_n \to x$ There is a question (could be from Ruddin's real analysis text) 

To show that $\lim\limits_{n \to \infty} f_n(x_n) = f(x)$, is equivalent to show that:
$$|f_n(x_n) - f(x)| < \epsilon$$
Which of the following approach is the correct way to show the above?


*

*$|f_n(x_n) - f(x)| \leq |f_n(x_n) - f_n(x)| + |f_n(x) - f(x)|$


We know $|f_n(x) - f(x)| < \epsilon$ since $f$ converges uniformly, we know $|f_n(x_n) - f_n(x)| < \epsilon$ because continuous function preserves convergence, hence $|f_n(x_n) - f(x)| < \epsilon$
or 


*$|f_n(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)|$


We have $|f_n(x_n) - f(x_n)| < \epsilon$ by uniform convergence, and $|f(x_n) - f(x)| < \epsilon$ by continuity, thus $|f_n(x_n) - f(x)| < \epsilon$
Which of the above method is the right way to prove the problem?
 A: The second approach is correct as mentioned in the comments.
There is a flaw in the argument of the first approach.  
That $|f_n(x) - f(x)| < \epsilon$ for all $x \in E$ when $n$ is sufficiently large is clear.  However you need to show that there exists $N(\epsilon,x) \in \mathbb{N},$ depending only on $\epsilon$ and $x$, such that if $n > N(\epsilon,x)$ we have $|f_n(x_n) - f_n(x)| < \epsilon$. Continuity of $f_n$ is not sufficient at this point.  You only can argue that for fixed $m$ we have $f_m(x_n) \to f_m(x)$ as $n \to \infty$.
If the sequence $(f_n)$ is equicontinuous then there exists $\delta(\epsilon)$ such that $|f_n(y) - f_n(x)| < \epsilon$ for all $n$ when $|y-x| < \delta(\epsilon)$.  Since $x_n \to x$, we can choose $N(\epsilon,x)$ such that $|x_n - x| < \delta(\epsilon)$ for $n > N(\epsilon,x)$ it follows that $|f_n(x_n) - f_n(x)| < \epsilon$.
As it turns out the uniform convergence of $f_n$ on $E$ -- even if it is not compact -- guarantees equicontinuity, but this is harder to prove than the proposition in the problem using your second approach. 
Addendum
Here is a specific example to show that you cannot argue that $|f_n(x_n) - f_n(x)| \to 0$ using only that $f_n$ is continuous and "... continuous function preserves convergence."
Consider the sequence of continuous functions where $f_n(x) = x^n$ on $[0,1]$.  Suppose $x_n = 1 - 1/n.$ 
Then $x_n \to 1$, but
$$\lim_{n \to \infty}|f_n(x_n) - f_n(1)| = \lim_{n \to \infty}|(1 - 1/n)^n - 1| = |e^{-1} - 1| \neq 0.$$
In this case, of course, the sequence of functions is not uniformly convergent.
