Looking for a lemma and its proof I recently acquired this following lemma for general topology and would like to know where (i.e. book) this was from, also if possible, a proof for it. Does this look familiar to anyone?

Lemma: Let $p: X \rightarrow Y$ be a closed map, then
(1): If $p^{-1} (y) \subset U$, where $U$ is an open subspace of $X$, then $p^{-1} (W) \subset U$, for some neighborhood $W \subset Y$ of $Y$.
(2): If $p^{-1} (B) \subset U$, for some subspace $B$ of $Y$ and some open subspace $U$ of $X$, then $p^{-1} (W) \subset U$, for some neighborhood $W \subset Y$ of $B$.

Thanks in advance.
 A: Well, this is a classical characterisation of closed maps, I don't know who proved it first.
Note that it is a sort of reverse continuity: instead of starting with a neighbourhood of $y = p(x)$ in the image and making sure $p$ maps a small enough neighbourhood of $x$ into it, we start with a neighbourhood of the inverse image (fibre) and get an open set in the image with reverse image inside it. It's a useful fact, and in fact, if a map $p$ satisfies it, it is a closed map (so we go back as well). 
2) implies 1), clearly (we take $B = \{y\}$), so we only need to prove 2).
So assume $p^{-1}[B] \subset U$, where $U \subset X$ is open, and $B \subset Y$.
Then $X \setminus U$ is closed, so as $p$ is a closed map, $p[X \setminus U]$ is closed in $Y$, and so we can define $W = Y \setminus p[X \setminus U]$, which is then open.
Now one has to verify two things: 
$B \subset W$: Let $y \in B$, arbitrary. Suppose $y \in p[X \setminus U]$, then $y = p(x)$ for some $x \in X \setminus U$. But then $x \in p^{-1}[B]$, as $p(x) = y \in B$ but not in $U$, contradicting $p^{-1}[B] \subset U$. So $y \in Y \setminus p[X \setminus U] = W$ as required.
$p^{-1}[W] \subset U$: let $x \in p^{-1}[W]$, so $p(x) \in W = Y \setminus p[X \setminus U]$. Suppose $x \notin U$, then $x \in X \setminus U$, so $p(x) \in p[X \setminus U]$. This is a contradiction: $p(x)$ cannot be in a set and its complement... So $x \in U$ and we are done.
