Rather than give formulas, I will describe the graphs of the functions, which are piecewise linear. Think of a “discretely moving wave”, where after the wave is done moving, we switch to a “thinner” wave.
Let $f_1$ go from $(0,1)$ to $(\frac{1}{2},0)$ to $(1,0)$. Let $f_2$ go from $(0,0)$ to $\frac{1}{2},1)$ to $(1,0)$. Let $f_3$ go from $(0,0)$ to $\frac{1}{2},0)$ to $(1,1)$.
Then let $f_4$ go from $(0,1)$ to $(\frac{1}{4},0)$ to $(1,0)$. Let $f_5$ go from $(0,0)$ to $(\frac{1}{4},1)$ to $(\frac{1}{2},0)$ to $(1,0)$. Then $f_6$ from $(0,0)$ to $(\frac{1}{4},0)$ to $(\frac{1}{2},1)$ to $(\frac{3}{4},0)$ to $(1,0)$. Then $f_7$ go from $(0,0)$ to $(\frac{1}{2},0)$ to $(\frac{3}{4},1)$ to $(1,0)$. Then $f_8$ from $(0,0)$ to $(\frac{3}{4},0)$ to $(1,1)$.
Next do the same thing with the partition $0\lt \frac{1}{8}\lt\frac{1}{4}\lt\frac{3}{8}\lt \frac{1}{2}\lt \frac{5}{8}\lt\frac{3}{4}\lt\frac{7}{8}\lt 1$. Then partitioning $[0,1]$ into $16$ equal subintervals, etc.
Each new batch of functions has a smaller integral, all positive, but converging to $0$. The functions are all continuous. And $f_n(x)$ does not converge for any $x$ because you can always find arbitrarily large values of $n$ where $f_n(x)=0$ and values where $f_n(x)$ is very close to $1$ (by approximating $x$ with a rational with denominator a power of $2$).
Best Answer
Your approach is correct and indeed $f_{n}\to x^{2}$ pointwise on $[0,1]$.
The definition of pointwise convergence of a sequence of functions $(f_{n}(x))_{n\in \mathbb{N}}$ on a set $I\subseteq \mathbb{R}$ says, $f_{n}\to f$ pointwise on $I$ if, and only if, $$(\color{red}{\forall x\in I})(\forall \varepsilon >0)(\color{red}{\exists N\in \mathbb{N}})(\forall n\in \mathbb{N})(n>N\implies |f_{n}(x)-f(x)|<\varepsilon.$$
Let the sequence of functions $f_{n}(x)=\left(x-\frac{1}{n}\right)^{2}$ and let's to show that $f_{n}\to f$ pointwise on $[0,1]$, with $f(x)=x^{2}$.
Let $x\in [0,1]$ and let $\varepsilon >0$. By the Archimedean property there exists a natural number $N>\frac{\varepsilon}{3}$. If $n>N$ we have \begin{align*} |f_{n}(x)-f(x)|&=\left|\left(x-\frac{1}{n}\right)^{2}-x^{2}\right|\\&=\left|x^{2}-\frac{2x}{n}+\frac{1}{n^{2}}-x^{2}\right|\\ &\leqslant \frac{|2x|}{n}+\frac{1}{n^{2}},\quad 0\leqslant x\leqslant 1,\\&\leqslant \frac{2}{n}+\frac{1}{n^{2}}\\&\leqslant \frac{3}{n},\\&<3\left(\frac{\varepsilon}{3}\right),\\&=\varepsilon. \end{align*} Therefore $f_{n}\to f$ pointwise on $[0,1]$.
Moreover $f_{n}\rightrightarrows f$ uniformly on $[0,1]$.
The definition of uniform convegence of a sequence of functions $(f_{n}(x))_{n\in \mathbb{N}}$ on a set $I\subseteq \mathbb{R}$ says, $f_{n}\rightrightarrows f$ uniformly on $I$ if, and only if, $$(\forall \varepsilon >0)(\color{red}{\exists N\in \mathbb{N}})(\forall n\in \mathbb{N})(\color{red}{\forall x\in I})(n>N\implies |f_{n}(x)-f(x)|<\varepsilon.$$
Let the sequence of functions $f_{n}(x)=\left(x-\frac{1}{n}\right)^{2}$ and let's to show that $f_{n}\rightrightarrows f$ uniformly on $[0,1]$, with $f(x)=x^{2}$.
Let $\varepsilon >0$, thus choose $N\in\mathbb{N}$ such that $\frac{1}{N^{2}}<\frac{\varepsilon}{2}$ and $\frac{2x}{N}<\frac{\varepsilon}{2}$. Thus for all $n>N$ we have \begin{align*} |f_{n}(x)-f(x)|&=\left|\left(x-\frac{1}{n}\right)^{2}-x^{2}\right|,\\&=\left|-\frac{2x}{n}+\frac{1}{n^{2}}\right|,\\&\leqslant \frac{2x}{n}+\frac{1}{n^{2}},\\&<\frac{2x}{N}+\frac{1}{N^{2}},\\&=\frac{\varepsilon}{2}+\frac{\varepsilon}{2},\\&=\varepsilon. \end{align*}
Therefore $f_{n}\rightrightarrows f$ uniformly on $[0,1]$.
Alternatively notice that $$\sup_{x\in [0,1]}\left|f_{n}(x)-f(x)\right|\to 0,\quad n\to +\infty$$ Therefore $f_{n}\rightrightarrows f$ uniformly on $[0,1]$. Then also $(f_{n}(x))_{n\in \mathbb{N}}$ is uniform Cauchy.