I saw the title statement in a Chinese translation of a Russian book, and I put doubt on it, so it's impossible to find companions or solutions. Since it's quite weird, and I can think of an example that may or may not be a counterexample (I can't prove the validity of it).
Imagine the sequence of all rationals: $r_1,r_2,...,r_n,....$ Now construct $f_n(x):=1$ at $r_1,r_2,...,r_n$, and have a V shape between $2$ successive $r$'s arranged from small to large. For instance, take $r_1=0,r_1=1,r_2=1/2,$ then my construction is $$f_2(x)=\begin{cases}|4x-1| & x\in[0,1/2],\\|4x-3| & x\in[1/2,1].\end{cases}$$ Then we take $F_n(x)=f_n(x)^n.$ I'm expecting $f_n$ to run all over the rationals, and results in 1's in the limit of $F_n,$ while the irrationals are less than $1$ and raising to the power of $n$ makes it go to zero.
But wait, I can't guarantee that the $f_n(\text{irrational})$ is less than a number that is strictly less than one, i.e.: I can't prove that the $F_n(\text{irrational})$ tends to zero as $n\to\infty.$
Can you prove or disprove the statement in the title? Thanks.
As Daniel points out, there is a general fact, but let me try to give a bit more details on a possible proof. Let $C_N = \{x\in[a,b]:|f_n|\leq N\textrm{ for all }n\}$. Since all $f_n$ are continuous, each set $C_N$ is closed. Then as $\sup_n|f_n(x)|$ is finite for any $x$, the interval $[a,b]$ can be written as the union of all $C_N$ as $N$ runs over all natural numbers. Now Baire's theorem says that at least one of $C_N$ contains an open interval. This means that there is an open interval on which the family $\{f_n\}$ is uniformly bounded. I leave the rest as an exercise.
Note that the above argument is usually called the uniform boundedness principle, because it derives uniform boundedness from pointwise boundedness. Let us recall Baire's theorem here.
Baire's theorem. Let $\{C_N\}$ be a countable collection of closed subsets of $[a,b]$ such that $\bigcup_N C_N=[a,b]$. Then at least one of $C_N$ contains a nontrivial interval, i.e., there exist $N$, $x\in [a,b]$, and $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon)\subset C_N$.
Proof. Suppose that $C_N$ does not contain any open interval, for any $N$. This means that any open interval $I$ in $[a,b]$ contains a point from $[a,b]\setminus C_N$, and so $I\cap([a,b]\setminus C_N)$ contains a nontrivial closed interval, because $[a,b]\setminus C_N$ is open (relative to $[a,b]$). Applying this with $I=(x-\varepsilon,x+\varepsilon)$ for some $\varepsilon<1$, we obtain $x_1\in [a,b]$ and $\varepsilon_1\in(0,1)$ such that $[x_1-\varepsilon_1,x_1+\varepsilon_1]\subset [a,b]\setminus C_1$. Similarly, there are $x_2\in [a,b]$ and $\varepsilon_2\in(0, \frac12)$ such that $$ [x_2-\varepsilon_2,x_2+\varepsilon_2]\subset (x_1-\varepsilon_1,x_1+\varepsilon_1) \cap ([a,b]\setminus C_2), $$ and so on, we get a sequence of intervals $(x_k-\varepsilon_k,x_k+\varepsilon_k)$ such that $\varepsilon_k\in(0, \frac1k)$ and $$ [x_k-\varepsilon_k,x_k+\varepsilon_k]\subset (x_{k-1}-\varepsilon_{k-1},x_{k-1}+\varepsilon_{k-1}) \cap ([a,b]\setminus C_k). $$ In particular, we have $x_k\in [x_n-\varepsilon_n,x_n+\varepsilon_n]$ for $k>n$, hence $\{x_k\}$ is Cauchy, and by completeness, there is $x\in [a,b]$ such that $x_k\to x$. By closedness, we have $x\in [x_n-\varepsilon_n,x_n+\varepsilon_n]$ for all $n$, and since $[x_n-\varepsilon_n,x_n+\varepsilon_n]\subset [a,b]\setminus C_n$, we have shown that there is $x\in [a,b]$ such that $x\not\in C_n$ for all $n$.