This examples (example 7.21, page 156) proves that the given sequence of functions has no uniformly convergent subsequence:
Let $$f_{n}(x)=\frac{x^2}{x^2+(1-nx)^2},\quad (0\leq x\leq 1, n=1,2,3,\dots)$$
Then $|f_{n}(x)|\leq 1$, so $f_{n}$ is uniformly bounded on $[0,1]$. Also $$\lim_{n\to\infty}f_n(x) =0\quad (0\le x \le 1)$$
but $$f_{n}\left(\frac 1n\right)=1 \quad(n=1,2,3,\ldots)$$
so that no subsequence can converge uniformly on $[0,1]$.
I have two questions. Is the claim that no subsequence can converge uniformly on $[0,1]$ justified because fixing $n$, we cannot estimate $f_{n}(x)$ at $1/n$ with $\epsilon$ precision and hence it doesn't converge for all $n$. So, any subsequence of it won't converge to the limit.
Also is the limit mentioned in the example $lim _{n}\to\infty=0$ referring to pointwise limit? Another question is why is it necessary to show that the given sequence of functions is bounded uniformly by $1$?
The example of Rudin is showing that Bolzano-Weierstrass theorem is not true for
continuous functions in a closed interval with the topology of uniform convergence. For real numbers, the B-W theorem says that every bounded sequence of real numbers has a convergent subsequence.
Then, the analogous question in this case would be: has every uniformly bounded sequence of continuous functions in $[0,1]$ a uniformly convergent subsequence?
And the answer is no. First, the sequence of the example is uniformly bounded by $1$ (it satisfies the hypothesis) but, there doesn't exist a subsequence $f_{n'}$ and a continuous function $f$ such that $$ \lim_{n' \rightarrow \infty} \sup_{x \in [0,1]} \vert f_{n'}(x) - f(x) \vert = 0 $$
If such a $f$ does exist, it must be $f \equiv 0$, the pointwise limit of the sequence, because, for every $x_0 \in [0,1]$, $$ 0 = \lim_{n' \rightarrow \infty} \vert f_{n'}(x_0) - 0\vert \leq \lim_{n' \rightarrow \infty} \sup_{x \in [0,1]} \vert f_{n'}(x) - f(x) \vert = 0 $$ However, $0$ is not a uniform limit for any subsequence, because $$ \sup_{x \in [0,1]} \vert f_{n}(x)\vert \geq f_n(1/n) = 1 > 0, $$ for all $n$, and this finishes the argument. The result that establishes a true analog for B-W theorem in this topology is Arzelà-Ascoli Theorem.