Prove that if $f:X\longrightarrow Y$ is closed and continuous then $Y$ is paracompact (and hausdorff) provided that $X$ is. I was studying paracompactness by Munkres topology text when I ran into the following exercise

Prove that paracompactness is an invariant with respect perfect maps defined in hausdorff spaces.

which I was not able to solve so that I search in solution in other topology text and I consulted the Engelikng topology text discovering that paracompactness is actually an invariant with respect closed maps so that I started to read the following proof he gave

but unfortunately there are some aspect I did not understand so that I thought to put a specific question where I ask clarification about: previously I point out that Engelking requires that paracompact spaces are Harsdorff whereas Munkres did not this so that with respect Munkres formalism the question became to prove that if $f$ is a closed and continuous map form a space $X$ to a space $Y$ then $Y$ is paracompact and hausdorff when $X$ is it. So exactly I did not understand the following things.

*

*Clearly by the lemma 5.1.6 the collection $\cal F_1$ satisfies $(11)$ but I do not really see why it must satisfies $(12)$ too: perhaps do it satisfies it vacuously ? However in this case could be better define the collection $\mathcal F:=\big\{\mathcal F_i\big\}$ for $i\ge 0$ instead that for $i>1$? Moreover since I am studying by Munkres can I use the following Munkres result instead that Engelking lemma above mentioned? indeed $U_s$ and $W_{s,k}$ are open and cover $X$, right?

*Clearly if $F_{s,k}$ is contained in $W_{s,k}$ then it is contained in $f^{-1}[U_s]$ because $W_{s,k}$ is there contained by construction; moreover we observe that
$$
W_{s,k}=f^{-1}[U_s\setminus f[E_{s,k-1}]
$$
or rather
$$
f[F_{s,k}]\subseteq f[W_{s,k}]\subseteq U_s\setminus f[E_{s,k-1}]
$$
which prove that $F_{s,k}$ is disjoint from $f[E_{s,k-1}]$. Did I explain well this?

*More precisely $V_{i,s}$ open because $\mathcal F_i$ and $\mathcal F_{i+1}$ are locally finite and because $f$ is a closed map, right?

*I do not really understand why $\cal V$ cover $Y$: moreover how take $i(y)$ such that $y\in F_{s(y),i(y)-1}$ if $s(y)$ is the minimum for $i(y)$? again why $y\notin F_{s,i(y)+1}$ whenever $s>s(y)$? So I courteously ask to explain very well why $\cal V$ cover $Y$ because actually it is the unique thing I did not truly understand.

*$U$ is open by analogous arguments given in $3$ here, right?

*Now $t(y)$ is the smallest element such that $y\in f[F_{y(y),i}]$ so that $y\notin\bigcup_{s<t(y)}F_{s,i}$ or rather
$$
y\in Y\setminus f\Biggl[\bigcup_{s<t(y)}F_{s,i}\Biggl]=M
$$
Then by $(12)$ we know that $f[F_{s,i+1}]$ is disjoint form $E_{s,i}$ with $s>t(y)$ and so from $f[F_{t(y),i}]$ so that if $y\in f[F_{y(y),i}]$ then $y\notin f[F_{s,i+1}]$ or rather
$$
y\in Y\setminus f\Biggl[\bigcup_{s>t(y)}F_{s,i+1}\Biggl]=N
$$
by arbitrariness of $s>t(y)$. So observing that
$$
U=(Y\setminus M)\cap(Y\setminus N)
$$
we conclude that $y$ is in $U$. I did explain well this?

*Finally if $V_{s,i}\subseteq f[F_{s,i}]$ for any $s\in\cal S$ then in particular it holds for $s<t(y)$ but we know that
$$
U\subseteq Y\setminus f\Biggl[\bigcup_{s<t(y)}F_{s,i}\Biggl]=\bigcap_{s<t(y)}\Big(Y\setminus f[F_{s,i}]\Big)
$$
so that $V_{s,i}$ is disjoint from $U$ if $s<t(y)$; analogously we know that $V_{s,i}\subseteq f\Big[\bigcup_{t\ge s} F_{t,i+1}\Big]$ for any $s\in\cal S$ and so for any $s>t(y)$ or rather
$$
V_{s,i}\subseteq f\Biggl[\bigcup_{t>s(y)}F_{t,i+1}\Biggl]
$$
but we also know that
$$
U\subseteq Y\setminus f\Biggl[\bigcup_{t>t(y)}F_{t,i+1}\Biggl]
$$
and thus we conclude that $V_{s,i}$ and $U$ are disjoint for $s>t(y)$: this prove that $U$ intersect at most $V_{t(y),i}$. So did I well explain this?

So could someone help me, please?
 A: For (4), let $y\in Y$. For each $i\in\Bbb Z^+$, $\mathscr{F}_i$ covers $X$, so there is a $<$-least $s(y,i)\in S$ such that $y\in f[F_{s(y,i),i}]$; let $s(y)=\min\limits_<\{s(y,i):i\in\Bbb Z^+\}$. There is then some $j\in\Bbb Z^+$ such that $s(y)=s(y,j)$, so that $y\in f[F_{s(y),j}]$, and we let $i(y)=j+1$, so that $y\in f[F_{s(y),i(y)-1}]$.
Let $s\in S$, and suppose first that $s(y)<s$; clearly
$$y\in f[F_{s(y),i(y)-1}]\subseteq f\left[\bigcup_{t<s}f[F_{t,i(y)-1}]\right]=f[E_{s,i(y)-1}]\,,$$
and from $(12)$ we know that $f[F_{s,i(y)}]\cap f[E_{s,i(y)-1}]$, so $y\notin f[F_{s,i(y)}]$. Now suppose that $s<s(y)$; the minimality of $s(y)$ implies that $s<s\big(y,i(y)\big)$, so $y\notin f[F_{s,i(y)}]$. Similarly, $s<s\big(y,i(y)+1\big)$, so $y\notin f[F_{s,i(y)+1}]$. Thus,
$$y\in Y\setminus\left(\bigcup_{\substack{s\in S\\s\ne s(y)}}f[F_{s,i(y)}]\cup\bigcup_{\substack{s\in S\\s<s(y)}}f[F_{s,i(y)+1}]\right)=V_{s(y),i(y)}\,,$$
and $\mathscr{V}$ is therefore an open cover of $Y$.
As for (1), since $(12)$ is asserted only for $i>1$, it imposes no requirement on $\mathscr{F}_1$, so yes, one could say that $\mathscr{F}_1$ satisfies it vacuously. In any case, a close examination of the argument shows that $(12)$ is applied only to $i(y)$, which is $j+1$ for some $j\in\Bbb Z^+$, so $i(y)>1$ anyway.
Your other points are okay, apart from a few typographical errors.
