How can pointwise boundedness on an interval prove uniform boundedness on a sub-interval? Edit:
I'm trying to use contradiction.  Suppose on a compact subinterval [c,d] $f_n(x)$ is not uniformly bounded over all x and for all n=1,2,...
Now cut the interval in half and look at the subinterval [d,e]...
If we do not get a uniform bound on this subinterval, cut in half again and repeat the procedure...
Eventually we would reach a singleton ...
Hmm... 
Assume that $f_n(x)$ is continuous over the interval [a,b], and that for each x in [a,b], ${f_n(x)}$ is bounded. Prove that there is a subinterval in  [a,b] such that $f_n(x)$ is uniformly bounded on this subinterval.
Any hints on how to get started on this problem are welcome.
Thanks,
 A: A Baire category argument would be recommended here, rather than a compactness argument like the nested interval attack.
For each integer $m=1,2,3, ...$ define the set
$$E_m = \bigcap_{n=1}^\infty \{x \in [a,b]: |f_n(x)| \leq m\}.$$
Since each $f_n$ is continuous, each set $E_m$ is closed.  Also, since each sequence $\{f_n(t)\}$ is bounded for a fixed $t\in [a,b]$ this means that
every point of $[a,b]$ is in one of the sets $E_1$, $E_2$, $E_3$, ...
No interval can be expressed as the union of a sequence of closed nowhere dense sets (thankyou Baire) so one of the sets $E_M$ (say) contains an interval $[c,d]$.  On that interval $|f_n(x)|\leq M$ for all $n$.

Example.  Here is an example to show that continuity is indeed needed here.  Let $\{r_n\}$ be an enumeration of the rationals in $[0,1]$.  For each $n=1,2,3, ...$ let $f_n(r_n)=n$ and $f_n(x)=0$ if $x\not=r_n$.  Each function $f_n$ is bounded but has a single point of discontinuity.  Certainly $\{f_n(t)\}$ is bounded for each fixed $t\in [0,1]$.  There is, however, no subinterval of $[0,1]$ on which $\{f_n\}$ is uniformly bounded since any such interval  contains infinitely many rationals.
A: What do we know?  Well, each $f_{n}$ is continuous on $[a,b]$, and thus bounded.
Also, for each $x \in [a,b]$, the sequence of real numbers $\{|f_{n}(x)| \}_{n = 1}^{\infty}$ is bounded by some $M_{x} > 0$ (in fact, let's let $M_{x}$ be the smallest such bound).
Now, we can't say that this implies $\{f_{n}\}$ is uniformly bounded on $[a,b]$, because the $M_{x}$'s could get arbitrarily large.
But we need to prove that on some sub-interval, the $M_{x}$'s can't get arbitrarily large, i.e., that for some sub-interval $[c,d]$, $\sup \limits_{x \in [c,d]} M_{x} < \infty$.  In this case, we could then let $M = \sup \limits_{x \in [c,d]} M_{x}$ and so we would have $|f_{n}| \leq M$ for all $n$ and $x \in [c,d]$, i.e., we would have a uniform bound.
$\text{ }$

Let's prove this by contradiction.  Suppose for each sub-interval $[c,d]$, $\sup \limits_{x \in [c,d]} M_{x} \not < \infty$.  Then if we consider the midpoint of $[a,b]$, which is $(a + b)/2$, we have that:
For each $n$, if $A_{n} = [\frac{a + b}{2} - \frac{1}{n}, \frac{a + b}{2} + \frac{1}{n}]$, then $\sup \limits_{x \in A_{n}} M_{x} \not < \infty$.
In particular, find $x_{n} \in A_{n}$ so that $M_{x_{n}} > n$.  Then by construction of $x_{n}$, we have $x_{n} \to (a+b)/2$, while $M_{x_{n}} \to \infty$.  
Since $M_{x_{n}}$ is the smallest bound of $\{|f_{m}(x_{n})|\}_{m = 1}^{\infty}$, this implies that $\{ |f_{n}( \frac{a + b}{2})| \}_{n = 1}^{\infty}$ is unbounded, which contradicts that the sequence $f_{n}$ is pointwise bounded.
