Assume that $f:\,\mathbb{R}\rightarrow\mathbb{R}$ is a continuous function. Assume also that exists a $M>0$ such that, for all $x \in \mathbb{R}$, we have $$\left|f\left(x\right)\right|\leq M.$$ Suppose that exists a sequence of function $f_n$ such that $f_n \rightarrow f$ uniformly. I want to prove that, for all $x \in \mathbb{R}$, we have $$\left|f_{n}\left(x\right)\right|\leq M$$ for every sufficiently large $n$.
My attempt. Since $f_n \rightarrow f$ uniformly for all $\epsilon>0$ exists a $N>0$ such that for all $x \in \mathbb{R}$ and $n \geq N$ we have $$\left|f_{n}\left(x\right)-f\left(x\right)\right|<\epsilon.$$ So $$\left|f_{n}\left(x\right)\right|<\left|f\left(x\right)\right|+\left|f_{n}\left(x\right)-f\left(x\right)\right|$$ $$\leq M + \epsilon.$$ Is this proof correct? Thank you.
Your considerations are correct, but they do not show that $|f_n(x)| \le M$ for all $x$ and all sufficiently large $n.$
You can not prove, what you want to prove!
Exanple : $f(x)=\frac{1}{1+x^2}$, then $\max \{|f(x)|: x \in \mathbb R\}=1=f(0)$.
Now take $f_n(x):=f(x)+\frac{1}{n}$. Then
$\max \{|f_n(x)|: x \in \mathbb R\}=1+\frac{1}{n}=f_n(0)$.