Can there be an unbounded sequence of equicontinuous functions? I am trying to find a equicontinuous sequence of functions $f_n$ on $(a, b)$ that is bounded somewhere but not everywhere.
I am thinking along the lines of 
$$f_n=\frac{1}{nx}$$ on $(0, 1)$, but this is obviously not equicontinuous.
Any hints?

I understand/interpret your question as follows: You want to find
an equicontinuous sequence $\left(f_{n}\right)_{n\in\mathbb{N}}$
of functions $f_{n}:\left(0,1\right)\to\mathbb{R}$ such that the
set
$$
U:=\left\{ x\in\left(0,1\right)\,\middle|\,\exists M_{x}\in\left(0,\infty\right)\,\forall n\in\mathbb{N}:\,\left|f_{n}\left(x\right)\right|\leq M_{x}\right\} 
$$
is not empty, but also not all of $\left(0,1\right)$.
As we will see now, such a sequence does not exist. To see this, we
will show that $U$ is both open and closed in $\left(0,1\right)$.
By connectedness of $\left(0,1\right)$, this implies $U\in\left\{ \emptyset,\left(0,1\right)\right\} $.


*

*$U$ is open: Let $x_{0}\in U$ be arbitrary. By equicontinuity,
there is $\varepsilon>0$ (potentially depending on $x_{0}$) with
$\left(x_{0}-\varepsilon,x_{0}+\varepsilon\right)\subset U$ and with
$\left|f_{n}\left(x_{0}\right)-f_{n}\left(y\right)\right|<1$ for
all $y\in\left(x_{0}-\varepsilon,x_{0}+\varepsilon\right)$. But this
implies
$$
\left|f_{n}\left(y\right)\right|\leq\left|f_{n}\left(x_{0}\right)\right|+1\leq M_{x_{0}}+1
$$
for all $n\in\mathbb{N}$, so that we get $\left(x_{0}-\varepsilon,x_{0}+\varepsilon\right)\subset U$.

*$U$ is closed in $\left(0,1\right)$. To see this, let $x_{0}\in\overline{U}\cap\left(0,1\right)$.
By equicontinuity, there is $\varepsilon>0$ with $\left(x_{0}-\varepsilon,x_{0}+\varepsilon\right)\subset U$
and with $\left|f_{n}\left(x_{0}\right)-f_{n}\left(y\right)\right|<1$
for all $y\in\left(x_{0}-\varepsilon,x_{0}+\varepsilon\right)$. But
since $x_{0}\in\overline{U}$, there is some $y\in U\cap\left(x_{0}-\varepsilon,x_{0}+\varepsilon\right)$.
This implies
$$
\left|f_{n}\left(x_{0}\right)\right|\leq\left|f_{n}\left(y\right)\right|+1\leq M_{y}+1
$$
for all $n\in\mathbb{N}$. Hence, $x_{0}\in U$.
