I have to show that if $f_n$ is a sequence of bounded functions that converges uniformly to $f$ on an interval I, then $f_n$ is uniformly bounded.
I understand what uniformly bounded means, there exists an $B$ such that for all x $\in I$ and for all $n\in N$ that the |$f_n$|
Should I start off with the assumption that $f_n$ is a sequence of bounded functions .... Therefore by definition for and $\epsilon$>0 that there exists an N such that if n>N, then |$f_n-f$|
You can use the fact that a uniformly convergent sequence $(f_n)$ is uniformly Cauchy; that is, given $\epsilon>0$, there is an $N$ so that for all $n, m\ge N$ and all $x$, one has $$ |f_n(x)-f_m(x)|<\epsilon\tag{1} $$ (one may prove this by using an "$\epsilon/2$-argument" applied to $f_n-f_m=f_n-f+f-f_m$, where $f$ is the limit function).
In your case, you can set $\epsilon=1$ and choose a fixed $N$ in accordance with equation $(1)$. It will then follow that for any $n\ge N$, $|f_n|$ is bounded by $M+1$, where $M$ is a bound for $|f_N|$ (you might furnish some details here).
So, we have a uniform bound, $M+1$, for the family $\{f_n:n\ge N\}$.
To wrap up, we need to find a uniform bound for all $f_n$.
Towards this end, if we can find a uniform bound, $L$, for the finite family $\{|f_1|,|f_2|,\cdots, |f_{N-1}|\}$, then $L+M+1$ will do the job.
I leave this final task for you...