If $x_n \rightarrow x \implies \lim_{n \rightarrow +\infty}$ $f_n (x_n) \rightarrow f(x)$ ,then $f_n \rightarrow f$ uniformly in each compact set $K $ I found this proof online and was having trouble understanding the last steps.

Lemma: Given that $f_n(x_n)$ converges to $f(x)$ , show that $f_n$ is uniformly convergent to $f$ on a compact set k.

Proof by contradiction:
Assume $f_n$ (on compact set K) does not converge uniformly to $f$. Then there exists a sequence of $x$'s and an increasing sequence of $n \epsilon N $ s.t.
$$| \; f_{2n_{k_r-1}}(x_k) \; - \; f_{2n_{k_r}}(x _k) \; | \geq \epsilon . $$
Then, by Bolzano Weierstrass (since K is compact) there is a convergent subsequence {$x_{k_r}$} that converges to $x$. 
Define a sequence $y_{n}$ so that
$$y_{n_r} = x\; \textrm{if} \; n \neq 2n_{k_r} \; \textrm{or if} \;   n \neq 2n_{k_r-1}, $$ 
and $$y_{2n_{k_r}} = y_{2n_{k_r-1}} = x_{k_r} .$$
Then $y_n$ conveges to $x$.
let {$Z_n$} = {$f_n(y_n)$} . This sequence is not cauchy since 
$$| \; z_{2n_{k_r-1}} \; - \; z_{2n_{k_r}} \; | \geq \epsilon.$$
Thus we have proved that if $f_n$ does not converge uniformly to $f$, 
$x_n \rightarrow x \implies \lim_{n \rightarrow +\infty}$ $f_n (x_n) \rightarrow f(x)$ is false 

My question is, couldn't we just use 
{$Z_n$} = {$f_n(x_{k_r}))$}?  Why did they define a new sequence that is potentially just {$x_{k_r}$}?
also, is all this $2n-1$, $2n$ notation necessary? could we use $n$ and $n+1$?
 A: If $f_n$ does not converge uniformly on some compact $K$:
There exists $r>0$ and a sub-sequence $(A(n))_n$ of $\Bbb N$ and a sequence $(x_{A(n)})_n$ in $K$, converging to $x\in K,$  such that $|f_{A(n)}(x_{A(n)})-f(x_{A(n)})|>r$ for all $n.$
Now there exists a  sub-sequence $(B(n))_n$ of $\Bbb N$ such that $B(n)>A(n)$ and $|f_{B(n)}(x_{A(n)})-f(x_{A(n)})|<r/2$ for all $n.$
We would like to have a sequence $Y=(y_n)_n$ in $K,$ converging to $x,$ whose $B(n)$-th entry $y_{B(n)}$  is $x_{A(n)} $ for every $n$. We would then have $f(x)=\lim_{n\to \infty}f_{B(n)}(y_{B(n)})=\lim_{n\to \infty}f_{B(n)}(x_{A(n)}).$
Then the sequences $S_1=(f_{A(n)}(x_{A(n)})_n$ and $S_2=(f_{B(n)}(y_{B(n)}))_n=(f_{B(n)}(x_{A(n)})_n$ both converge to $f(x).$ But for every $i$, the $i$-th term of $S_1$ and the $i$-th term of $S_2$ differ by more than $r/2,$ which makes it  impossible for both series to converge to $f(x).$ So we obtain the desired paradox.
To obtain $Y$ let $y_j=x_{A(1)}$ for $j\le B(1),$ and for $B(n)<j\le B(n+1)$ let $y_j=x_{A(n+1)}.$
Remarks. (1).  We could combine $S_1$ and $S_2$ into a single sequence before deriving the contradiction, which seems to be the style of the proof you found, but it's not necessary. (2). The converse to this result is also  true.
