Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of functions defined on $[a,b]$ Assume that each $f_n$ is bounded, so $|f_n| \le M_n$ for all $x \in [a,b]$. If $\{f_n\}_{n=1}^{\infty}$ converges uniformly to $f$, then $f$ must also be bounded.
So here is what I have.
Assume $\{f_n\}_{n=1}^{\infty}$ converges uniformly to $f$. Then given $\epsilon>0$, there exists an $N$ such if $n>N$, then $|f(x)-f_n(x)| < \epsilon$.
I am stuck on where to go from here. Should I use the fact that each $f_n$ is bounded to get a bound for $f$ in a sense?
You are getting there. Take $\epsilon = 1$. Then there exists $N$ with the property that $n \ge N$ implies that $|f_n(x) - f(x)| \le 1$ for all $x \in [a,b]$. In particular, $|f(x)| \le 1 + |f_N(x)| \le 1 + M_N$ for all $x \in [a,b]$. Thus $f$ is bounded.