Here is a prelim problem and my attempted solution, which seems a bit complicated for a prelim exercise, so I am wondering if there is a shorter proof: Let $X$ be a Baire Space, $(Y,d)$ a metric space, $(f_n)_{n\in \mathbb N}$ a sequence of continuous functions such that $(f_n(x))_{n\in \mathbb N}$ is Cauchy for each $x\in X.$ The claim is that in this case, $(f_n)_{n\in \mathbb N}$ is equicontinuous at some point $x\in X,$ i.e. there is an $x_0\in X,$ and an open set $V\in X$ such that for all $\epsilon>0,\ d(f_n(x),f_n(x_0))<\epsilon$ whenever $x\in V.$ and $n\in \mathbb N.$ The hint is to consider $F_{mp}=\left \{ x\in X:d(f_m(x),f_n(x))\le 1/p\ \text{for every}\ n\ge m \right \}.$ The $F_{mp}$ are closed because the $f_n$ are continuous.
$1).$ Since $(f_n(x))_{n\in \mathbb N}$ is Cauchy for each $x\in X,$ we have, for each $p\in \mathbb N,\ \bigcup_{m\in \mathbb N}F_{mp}=X.$ Since $X$ is Baire, at least one $F_{mp}$ has non-empty interior so $F^{\circ}_{mp} \neq \emptyset.$
$2).$ Using the definition of equicontinuity at a point, we have that every element of $S=\bigcap_{p\in \mathbb N}\bigcup_{m\in \mathbb N}F^{\circ}_{mp}$ is a point at which $(f_n(x))_{n\in \mathbb N}$ is equicontinuous.
We want to prove that $S\neq \emptyset, $ so towards a contradiction, assume that $S=\emptyset.$ Then,
$3).\ X=S^{c}=\left ( \bigcap_{p\in \mathbb N}\bigcup_{m\in \mathbb N}F^{\circ}_{mp} \right )^{c}=\bigcup_{p\in \mathbb N}\left ( \bigcup_{m\in \mathbb N}F^{\circ}_{mp} \right )^{c}=\bigcup_{p\in \mathbb N}\left ( \bigcap_{m\in \mathbb N}(F^{\circ}_{mp})^{c} \right ).$ Now,
$4).\ \left (F^{\circ}_{mp} \right )^{c}=\left \{ x\in X:d(f_m(x),f_n(x))>1/p\ \text{for some}\ n\ge m \right \}.$
So, it suffices to prove that $\left (F^{\circ}_{mp} \right )^{c}$ has empty interior - it is already closed - and for this, we mimic the "usual" Baire-Osgood argument:
Let $U$ be open in $X$. Then, $\bigcup_{m\in \mathbb N}(F_{mp}\cap U)=U,$ so since $U$ is a Baire Space in its own right, there is a $V$, open in $U$ such that $V\subseteq F_{mp}\cap U$ for some integer, $m.$ But since $U$ is open in $X,\ V$ is, too and since $V\subseteq F_{mp}$, we have that $d(f_m(x),f_n(x))\le 1/p$ whenever $n\ge m$. Since $V\subseteq U,$ we have shown that there are elements in $U$ not in $\left (F^{\circ}_{mp} \right )^{c}.$ And this finishes the proof.