Let ${f_n}$ be a sequence of real valued continuous functions converging pointwise on $\Bbb R$. Show that there exists a number $M>0$ and an interval $I \subset \Bbb R$ such that $\sup\{ |f_n(x)|:x \in I \} \le M$ for all $n$.
The idea here is to use the uniform boundedness principle or Baire category theorem, it's just a little unclear for me the best one to use, and the best way to apply it. I tried looking at the proof of the uniform boundedness principle, but for that, you need a Banach space $X$ and a normed linear space $Y$. I assume $Y$ is the real numbers, but what would I choose for $X$? Continuous functions on $\mathbb R$ under some norm making it complete?
Hint Suppose the the contrary. Then choose an arbitrary point in $\Bbb R$, call it $x_0$. Consider the open interval $I_1=B(x_0,1)$. Then there exists $x_1$ in such that $f_{n_1}(x_1)\geqslant 1$. Choose $\varepsilon_1$ such that $I_2= B(x_1,\varepsilon_1)\subsetneq I_2$.
If we had $|f_n(x)|\leqslant 2$ for any $n>n_1$ and any $x\in I_2\smallsetminus\{x_1\}$, the claim would be true. So there exists $x_2\neq x_1$ and $n_2>n_1$ such that $f_{n_2}(x_2)\geqslant 2$. We may take $n_1<n_2$ and $x_1\neq x_2$ for finiteness conditions: deleting a finite amount of points or indices cannot alter the fact the claim doesn't hold.
Continuing, we produce a sequence of distinct points $(x_n)_{n\geqslant 1}$, and a sequence of nested closed intervals $(\overline{I_n})_{n\geqslant 1}$ such that $x_i\in I_i$ $f_{n_i}(x_i)\geqslant i$. By Cantor's nested interval property, there exists $x\in \bigcap\limits_{i\in I} \overline{I_i}$. I claim that $y_n=f_n(x)$ cannot converge.
ADD Note the proof is pretty Baire-ish.
As suggested by Daniel Fischer (thought he'd post it) another solution would be to note that $$\Bbb R=\bigcup_{m\geqslant 0}\bigcap_{n\geqslant 1}\{x\in\Bbb R:|f_n(x)|\leqslant m\}$$ and directlly invoke Baire's theorem.