Here is Prob. 11, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
Let $f \colon X \rightarrow Y$ be continuous. Show that if $X$ is Lindelof, or if $X$ has a countable dense subset, then $f(X)$ satisfies the same condition.
My Attempt:
Let $X$ and $Y$ be topological spaces, and let $f \colon X \rightarrow Y$ be a continuous map.
Case 1. Suppose that $X$ is Lindelof.
Let $\mathscr{A}$ be an open covering of $f(X)$ regarded as a subspace of $Y$.
For each $V \in \mathscr{A}$, we can find an open set $V^\prime$ of $Y$ such that
$$ V = f(X) \cap V^\prime. \tag{0} $$
Let $\mathscr{A}^\prime$ be the open covering of $Y$ given by
$$ \mathscr{A}^\prime \colon= \left\{ \, V^\prime \, \colon \, V^\prime \mbox{ is open in $Y$ and } f(X) \cap V^\prime \in \mathscr{A} \, \right\}. \tag{Definition 0} $$
[Please refer to (0) above.]
Let $V \in \mathscr{A}$ and $V^\prime \in \mathscr{A}^\prime$ for which (0) above holds. Then we find that
$$
\begin{align}
f^{-1}(V) &= f^{-1} \left( f(X) \cap V^\prime \right) \\
&= f^{-1} \big( f(X) \big) \cap f^{-1} \left( V^\prime \right) \\
&= X \cap f^{-1} \left( V^\prime \right) \\
&= f^{-1} \left( V^\prime \right). \tag{1}
\end{align}
$$
Moreover, as $V^\prime$ is an open set of $Y$ and as the mapping $f \colon X \rightarrow Y$ is continuous, so the inverse image $f^{-1} \left( V^\prime \right) = f^{-1} (V)$ is an open set of $X$.
We note that
\begin{align}
f(X) &= \bigcup_{V \in \mathscr{A}} V \\
&= \bigcup_{V^\prime \in \mathscr{A}^\prime} \left( f(X)\cap V^\prime \right) \\
&= f(X) \cap \left( \bigcup_{V^\prime \in \mathscr{A}^\prime} V^\prime \right) \\
&\subset \bigcup_{V^\prime \in \mathscr{A}^\prime} V^\prime,
\end{align}
which implies that
$$
f(X) \subset \bigcup_{V^\prime \in \mathscr{A}^\prime} V^\prime.
$$
Now since
$$ \bigcup_{V \in \mathscr{A}} V = f(X) \subset \bigcup_{V^\prime \in \mathscr{A}^\prime} V^\prime, $$
therefore we obtain
\begin{align}
f^{-1} \left( \bigcup_{V \in \mathscr{A}} V \right) &= f^{-1}\big( f(X) \big) \\
&\subset f^{-1} \left( \bigcup_{V^\prime \in \mathscr{A}^\prime} V^\prime \right) \\
&\subset X,
\end{align}
[Of course all the inverse images are subsets of the domain.] which simplifies to
$$
\bigcup_{V \in \mathscr{A}} f^{-1} \left( V \right) = X = \bigcup_{V^\prime \in \mathscr{A}^\prime} f^{-1} \left( V^\prime \right). \tag{2}
$$
Thus the collection
$$
\mathscr{A}_X \colon= \left\{ \, f^{-1} (V) \, \colon \, V \in \mathscr{A} \, \right\} = \left\{ \, f^{-1} \left(V^\prime \right) \, \colon \, V^\prime \in \mathscr{A}^\prime \, \right\}
$$
is an open covering of the Lindelof space $X$, and therefore some countable subcollection of $\mathscr{A}_X$ also covers $X$; let one such countable subcollection be
$$
\left\{\, f^{-1} \left( V_n \right) \, \colon \, n \in \mathbb{N} \, \right\} = \left\{\, f^{-1} \left( V_n^\prime \right) \, \colon \, n \in \mathbb{N} \, \right\}.
$$
[Please refer to (1) above.]
Finally since
$$
X = \bigcup_{n \in \mathbb{N} } f^{-1} \left( V_n \right),
$$
therefore we obtain
\begin{align}
f(X) &= f \left( \bigcup_{n \in \mathbb{N} } f^{-1} \left( V_n \right) \right) \\
&= \bigcup_{n \in \mathbb{N} } f \left( f^{-1} \left( V_n \right) \right) \\
&\subset \bigcup_{n \in \mathbb{N}} V_n \\
&\subset f(X),
\end{align}
[The last inclusion follwos from the fact that the sets $V_n$ are in the covering $\mathscr{A}$ of $f(X)$.] and hence
$$
\bigcup_{n \in \mathbb{N} } V_n = f(X).
$$
Thus the collection
$$
\left\{ \, V_n \, \colon \, n \in \mathbb{N} \, \right\}
$$
is a countable subcollection of $\mathscr{A}$ that also covers $f(X)$.
This shows that every open covering $\mathscr{A}$ of $f(X)$ has a countable subcollection also covering $f(X)$.
Hence $f(X)$ is Lindelof (as a subspace of $Y$) whenever $X$ is a Lindelof space and $f \colon X \rightarrow Y$ is a continuous mapping.
Am I right?
Case 2. Next, suppose that $X$ is separable. Let $D$ be a countable dense subset of $X$. Then $D \subset X$ such that $\overline{D} = X$, and since $f \colon X \rightarrow Y$ is continuous, therefore by Theorem 18.1 (2) in Munkres we obtain
$$
f(X) = f\left( \overline{D} \right) \subset \overline{ f(D) },
$$
and hence by Theorem 17.4 in Munkres
$$
\left(\overline{f(D)}\right)_{\mbox{in } f(X)} = f(X) \cap \overline{f(D)} = f(X),
$$
that is,
$$
\left(\overline{f(D)}\right)_{\mbox{in } f(X)} = f(X). \tag{3}
$$
Here $\overline{f(D)}$ denotes the closure of $f(D)$ in the topological space $Y$.
Moreover, as $D$ is a countable subset of $X$ and as $f \colon X \rightarrow Y$ is a single-valued map, so we can conclude that $f(D)$ is also a countable subset of $f(X)$.
From (3) above and what has been stated in the preceding paragraphs, we can conclude that $f(X)$ has a countable dense subset $f(D)$ whenever $X$ has a countable dense subset $D$.
Hence $f(X)$ is separable (as a subspace of $Y$) whenever $X$ is separable and $f \colon X \rightarrow Y$ is continuous.
Am I right?
Are both parts of my proof correct? If so, are my presentations of both proofs also clearly enough understandable? Or, are there any issues with either proof?
Best Answer
Your reasoning is correct, but could be much shorter:
If $f$ is continuous, $f[\overline{D}] \subseteq \overline{f[D]}$ for any $D \subseteq X$. (this is somewhere in Munkres too).
If $D$ is dense, the left hand side is $f[X]$ and so $f[D]$ is dense in $f[X]$ immediately. If $D$ is countable, so is $f[D]$. That's all there is to it.
The Lindelöf proof is basically copy-paste from the image of compactness proof: open cover of $Y$ pulls back to open cover on $X$, and the countable subcover there tells us what sets form a subcover of the original cover. The only thing that changes is countable instead of finite.
IMHO, the proof should convey why the statement holds and focus less on pretty trivial details, or notational stuff, that should already be known to all readers anyway. Focus on the new idea, if any.