How to achieve uniform boundedness without functional analysis? If $f_n(x)$ is cts on [a,b], and pointwise-bounded on this set, how could I show that there must exist a subinterval such that we achieve a uniform bound?  I've spent awhile on this problem but still have no real progress on it unfortunately.
I've read online a bit today and achieving uniform boundedness is apparently one of the big theorems in functional analysis.  But I am preparing for advanced calculus / non-measure theoretic analysis exams -- the above question is a former exam question.
Do you see a way to solve it without any version of Baire category theorem or functional analysis?  
Contradiction seems to be the way to go - but I'm not sure whether to contradict the assumption of continuity, pointwise boundedness or perhaps both ... 
Thanks,
 A: The Baire theorem is the more natural way to your result, but perhaps what follows is an alternative:
Suppose that there do not exists such an interval. Put $u_0=a$, $v_0=b$. We can find a $x_0\in [a,b]$ and a $n_0$ such that $|f_{n_0}(x_0)|>1$. By continuity, we can find $u_1,v_1$, with $u_0<u_1<v_1<v_0$ and $\displaystyle v_1-u_1 \leq \frac{v_0-u_0}{2}$, such that $|f_{n_0}(x)|>1$ for all $x\in [u_1,v_1]$. Now we can find a $x_1\in [u_1,v_1]$ and a $n_1>n_0$ such that $|f_{n_1}(x_1)|>2$, and we can find $u_2,v_2$ such that $u_1<u_2<v_2<v_1$ and $\displaystyle v_2-u_2\leq  \frac{v_1-u_1}{2}$, etc. Hence by induction, we get a sequence of intervals $[u_k,v_k]$, with $u_k<u_{k+1}<v_{k+1}<v_k$ , $\displaystyle v_{k+1}-u_{k+1}\leq  \frac{v_k-u_k}{2}$, and $n_{k+1}>n_k$, with the property that $|f_{n_k}(x)|>k+1$ on $[u_k,v_k]$. Now $u_k$, $v_k$ converge to a same limite $x$, and we have $x\in [u_k,v_k]$ for all $k$, and hence $|f_{n_k}(x)|>k+1$ for all $k$, a contradiction. 
A: [We have continued from here where the same question was first raised by Lebron.]
Mr Kelenner's proof uses only the basics of continuity and the nested interval argument.  So, on the face of it, it might seem that Mr Lebron James (who is currently taking a well-deserved respite from his basketball career) should be pleased that he has avoided a more sophisticated argument.
However ... the Baire category theorem on the real line is not deep, sophisticated, or difficult, and should by no means be avoided.  A short proof is available using only the same nested interval argument.  Do understand there is no functional analysis here and no measure theory.  Just basic real analysis ideas provable by nested interval arguments. 
Indeed Mr  Kelenner's proof is just a category argument well-concealed.  The reason he was able to formulate the argument is that he already well knows the nature of the problem and can write it up in a way that contains anyway the essence of the Baire category proof.  So instead of appealing directly to the Baire theorem, he slips a proof into the solution without comment.

Here is the essence:  Suppose that $E_1$, $E_2$, ... is a sequence of
  closed sets covering $[a,b]$.  Then one of these sets contains an
  interval $[c,d]$.  If not, then choose a nested sequence of intervals
  $\{[a_n,b_n]\}$ with $[a_n,b_n]$ containing no point of $E_n$.   The
  intervals shrink to a point $z$ that does not belong to any of the
  sets $E_n$, which is a contradiction.

Now let me dissect the Kelenner  proof.  Define $A_m$ to 
be the collection of all points $x \in [a,b]$ such that
$|f_n(x)|\leq m$ for all $n$.  By continuity each $A_m$ is
closed and by the pointwise bounded assumption every point of $[a,b]$ is in one of the sets.   
If there is no subinterval contained in any
one of the set $A_m$ then  choose a nested sequence of intervals $[a_m,b_m]$ with $[a_m,b_m]$
containing no point of $A_m$.   
The intervals shrink to a point $z$ that does not belong to any of the sets $A_m$, which is a contradiction. 
[The Kelenner  write  up just directly finds the intervals outside the $A_m$ without commenting that $A_m$ is closed and uses continuity there.]
One can see that the proof already contains anyway a proof of the Baire category theorem.

Moral:  You want to avoid learning the simple Baire category theorem but, in the end, your proof has to include exactly the same reasoning that you wished to avoid.  Things get easier when you learn a wee bit more, not harder.  You have to learn this trick anyway, why not give it a name and use it whenever you need it.
Here are some more problems that you can try for practice that will illustrate.  Try proving using only the nested interval argument and then try for a category argument.  They are from
Elementary Real Analysis. 

Problem 6.9.3 Suppose that  $f_n:[a,b]\to\mathbb{R}$ are continuous and $\lim_{n\to\infty}f_n(x)=0$ for each $x$.  Show that,
  for any $\epsilon>0$, there is an interval $[c,d]$ so that $|f_n(x)|<
  \epsilon$ for all $x\in [c,d]$. (Show not necessarily the case that
  this holds on $[a,b]$.)
Problem 6.9.4 Suppose that  $f_n:[a,b]\to\mathbb{R}$ are continuous and $\lim_{n\to\infty}f_n(x)=\infty$ for each $x$.  Show
  that, for any $M>0$, there is an interval $[c,d]$ so that $f_n(x)>M$ for
  all $x\in [c,d]$. (Show not necessarily the case that this holds on
  $[a,b]$.)

