Consider
For $n\in \mathbb{N}$ and $k\in \{0, 1, 2, ..., 2^{n}-1 \}$ is defined
$$I_{k}^{n}=\left[\frac{k}{2^{n}}, \frac{k+1}{2^{n}}\right)$$
and $f_{n}:[0, 1) \rightarrow \mathbb{R}$ is defined by
$$f_{n}(x)=\sum_{k=0}^{2^{n}-1} \frac{k}{2^{n}}1_{I_{k}^{n}}(x)$$
Where for $A\subseteq \mathbb{R}$, the function $1_{A}$ is the characteristic of $A$ such that $1_{A}(x)=1$ if $x\in A$ and $1 _{A}(x)=0$ if $x\notin A$.
I want to determine if the given sequence of functions converges uniformly in $[0, 1)$. Let's analyze a little bit the graph of the succession, comparing with $f(x)+0.1$, $f(x)-0.1$ from where $f(x)=x$
It seems that the images of $f_{n}$ do not fit completely in the band created by the shifted functions; so it might seem that the function does not converge uniformly on $[0, 1)$.
One way to see that this is the case is assume uniform convergence and arrive at a contradiction.
If it's then for any $\epsilon >0$ we can find $n_{0} \in \mathbb{N}$ which depends on epsilon s.t. for all $n \geq n_{0}$ and all $x_{0} \in [0, 1)$ we have that
$$|f_{n}(x)-f(x)|<\epsilon$$
In particular $$|f_{n_{0}}(x)-f(x)|<\epsilon$$
I think that in order to build the contradiction I should find some $x_{0} \in [0, 1)$ and some $\epsilon>0$ s.t.
$$|f_{n_{0}}(x_{0})-f(x_{0})|\geq \epsilon$$
This is
$$ \left| \sum_{{k=0}}^{{2^{n_{0}}-1}} \frac{k}{2^{n_{0}}}1_{I_k^{n_{0}}}(x_0) - x_0 \right|=\left| \frac{k}{2^{n_{0}}} \right| \geq \epsilon. $$
We could choose $x_{0}=0$ and then $|f_{n_{0}}(x_{0})-f(x_{0})|=\dfrac{k}{2^{n_{0}}}$
But here what remaining choices would help me to arrive at the contradiction? Any suggestions would be appreciated!
Your initial guess is incorrect, these functions are essentially just step functions, where the step size gets smaller and smaller as the number of steps gets smaller and smaller, this is evident from the graphs, but lets try to prove it;
$\mathbf{Claim}$: $f_n(x)$ is a step function that takes $2^n$ steps of size $2^{-n}$ from $0$ to $1$.
I.e.
$$ f_n(x) = \frac{\left\lfloor 2^n x\right\rfloor }{2^n} $$
$proof$: let $x_0 \in [0,1)$. Then there is exactly one $k\in \{0,1,...,2^{n}-1\}$ such that
$$ x_0\in \left[ \frac{k}{2^n}, \frac{k+1}{2^n} \right) $$
Since these intervals form a disjoint cover.
Then let that $k$ be $k_0$. So then
$$ \delta_{k,k_0} = 1_{I^n_k}(x_0) = \begin{cases} 1 && \text{ if }k=k_0 \\ 0 && \text{otherwise} \end{cases} $$
($\delta_{k,k_0}$ called the kronecker-delta function, the discrete version of the dirac-delta, is simply the piecewise function depending on $k$ and $k_0$ defined in this way, here we are treating $x_0$ as a "constant" )
\begin{align} f_{n}(x_0) &=\sum_{k=0}^{2^{n}-1} \frac{k}{2^{n}}1_{I_{k}^{n}}(x_0) \\ &= \sum_{k=0}^{2^{n}-1} \frac{k}{2^{n}}\delta_{k,k_0} \\ \end{align}
And since all of the terms in the sum except the term corresponding to $k=k_0$ are $0$ this becomes
\begin{align} f_n(x_0) = \frac{k_0}{2^n} \end{align}
And if you carefully calculate what $k_0$ is for a given $x_0$ you'll find that
$$ k_0 = \left\lfloor 2^n x_0\right\rfloor $$
Since
$$ \frac{ \left\lfloor 2^n x_0 \right \rfloor}{2^n} \leq x_0 < \frac{\left\lfloor 2^n x_0 \right \rfloor + 1}{2^n} $$
So that
$$ f_n(x) = \frac{\left\lfloor 2^n x\right\rfloor }{2^n} $$
As desired. From here on, it should be fairly simple for you to prove that this function converges uniformly to
$$ f(x) = x $$
(Hint, think of $f_n$ as flooring $x$ to the neerest integer multiple of $\frac{1}{2^n}$ )
Edit: Some more explanation:
We want to show that we can make
$$ \left| \frac{\lfloor 2^n x\rfloor}{2^n} - x\right| $$
Arbitrarily smal for any $x$ by choosing a large enough $n$. Note that for any real number $a$,
$$ \lfloor a \rfloor \leq a < \lfloor a\rfloor + 1$$
So in particular
$$ \lfloor 2^n x \rfloor \leq 2^nx < \lfloor 2^n x\rfloor + 1$$
Then dividing through by $2^n$ we get
$$ \frac{\lfloor 2^n x \rfloor}{2^n} \leq x < \frac{\lfloor 2^n x\rfloor}{2^n} + \frac{1}{2^n} $$
What can we conclude from this?