A sequence of functions converges uniformly if and only if for each sequence $\{x_{n}\}$.... I'm stuck with a proof of the  theorem (from Real Analysis via Sequences and Series by Charles Little) :
Theorem 10.2.4 Let $\{f_{n}\}$ be a sequence of functions on a nonempty set $I$. Then the sequence converges uniformly to $f$ on $I$ if and only if
\begin{equation}
\lim_{n\rightarrow\infty}(f_{n}(x_{n})-f(x_{n}))=0
\end{equation}
for each sequence $\{x_{n}\}$ in $I$.
I have troubles in this sense: $\fbox{$\Longleftarrow$}$.
This is what I have: Suppose by contradiction that the sequence $\{f_{n}\}$ does not converges uniformly on I, so there is $\varepsilon>0$, such that 
for $1$ there is an integer $k_{1}\geq 1$ and $x_{k_{1}}\in I$ for which $|f_{k_{1}}(x_{k_{1}})-f(x_{k_{1}})|\geq\varepsilon$
$$\vdots$$
for $m$ there is an integer $k_{m}\geq m$ and $x_{k_{m}}\in I$ for which $|f_{k_{m}}(x_{k_{m}})-f(x_{k_{m}})|\geq\varepsilon$.
So I have a sub-sequence $\{f_{k_{n}}\}$ of $\{f_{n}\}$. Now, let $\{x_{n}\}$ be a sequence in $I$ having $\{x_{k_{m}}\}$ as a sub-sequence. Then the sub-sequence $\{|f_{k_{m}}(x_{k_{m}})-f(x_{k_{m}})|\}$ of $\{|f_{n}(x_{n})-f(x_{n})|\}$ does not converge to $0$, then $\{|f_{n}(x_{n})-f(x_{n})|\}$ does not converge to $0$. Which is a contradiction $\square$
It's my proof correct? The author  constructs the sub-sequence $\{f_{k_{n}}\}$ of $\{f_{n}\}$ by induction saying:
Suppose that positive integers $k_{1},\ldots k_{n}$ have been defined for some $n\in\mathbb{N}$, and that $x_{k_{j}}\in I$ and
\begin{equation}
|f_{k_{j}}(x_{k_{j}})-f(x_{k_{j}})|\geq\varepsilon
\end{equation}
for all $j$. Suppose also that $k_{j}<k_{j+1}$ for each $j<n$. Then there exist an integer $k_{n+1}\geq k_{n}+1$ and an $x_{k_{n+1}}\in I$ such that
\begin{equation}
|f_{k_{n+1}}(x_{k_{n+1}})-f(x_{k_{n+1}})|\geq\varepsilon
\end{equation}
Finally he says
We have now constructed a sub-sequence $\{f_{k_{n}}\}$ fo $\{f_{n}\}$ by induction. Let $\{x_{n}\}$ be a sequence in $I$ having $\{x_{k_{n}}\}$ as a sub-sequence. For instance, since $I\neq\emptyset$ we may choose $a\in I$ and set $x_{n}=a$ for each positive integer $n\not\in\{k_{1},k_{2},\ldots\}$. Then the sub-sequence $\{|f_{k_{m}}(x_{k_{m}})-f(x_{k_{m}})|\}$ of $\{|f_{n}(x_{n})-f(x_{n})|\}$ does not converge to $0$, and the proof is complete.
I understand his proof, but I  was wondering  if is really necessary construct that sub-sequence by induction? 
Thanks.
 A: Your construction is also essentially inductive (and essentially the same as his). You did miss a small point though - you need to choose $k_m>k_{m-1}$ or else you will not satisfy the definition of subsequence ($k_m$ needs to be an increasing sequence of natural numbers). This is the inductive part - you need the previous value to construct the next one. 
A: (I). (Remarks). Your proof is OK and is the same as in the book, except you should require $k_{m+1}>k_m.$ Otherwise you might end up with, for example, $k_m=k_1$ for all $m,$ and the conclusion $\neg (\lim_{n\to \infty}|f_n(x_n)-f(x_n)|=0)$ would not follow. It is crucial that $\{k_m:m\in \mathbb N\}$ is an infinite set. 
Also you should delete references to contradiction. You are proving  RHS$\implies$ LHS by a direct proof that ($\neg$ LHS)$\implies (\neg RHS$).
The name for this kind of construction is  recursion, not  induction. Although you could employ induction to define $k_{m+1}$ to be the LEAST $k>k_m$ such that there exists $x_k$ such that $|f_k(x_k)-f(x_k)|\geq  \epsilon$. 
My edit of your Q was to fix some trivial typos and to delete the mystery word "netx" from the first sentence. And to change "subsequence" to "sub-sequence" to make the computer feel better, as it red-lines "subsequence".
(2). Non-recursively:  For any real-valued functions $f,g$ on $I$ let $\|f-g\|=\min (1, \sup \{|f(x)-g(x)|:x\in I\}).$ Then uniform convergence of $(f_n)_n$  to $f$ is equivalent to $\lim_{n\to \infty}\|f_n-f\|=0.$ 
Suppose $(f_n)_n$ does not converge uniformly to $f.$  Then take $r\in (0,1) $ such that $S_r=\{n:\|f_n-f\|>r\}$ is an infinite set.
So for each $n\in S_r$ take $x_n\in I$ such that $|f_n(x_n)-f(x_n)|>r.$  For $n\in \mathbb N$ \ $S_r$ let $x_n$ be any member of $I$ that you want.  Then $$\neg (\lim_{n\to \infty}|f_n(x_n)-f(x_n)|=0).$$
