Here is Prob. 12, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
Let $f \colon X \rightarrow Y$ be a continuous open map. Show that if $X$ satisfies the first or the second countability axiom, then $f(X)$ satisfies the same axiom.
My Attempt:
Let $X$ and $Y$ be topological spaces, and let $f \colon X \rightarrow Y$ be a continuous open map.
Case 1. Suppose that $X$ is first-countable.
Let $y = f(x)$ be any point in $f(X)$. Then $x \in X$, and let $$ \left\{ B_{x, n} \colon n \in \mathbb{N} \right\} \tag{0} $$ be a countable local basis at $x$. Then each set $B_{x, n}$ is an open set of $X$ such that $x \in B_{x, n}$, and for any open set $U$ of $X$ such that $x \in U$, there exists a set $B_{x, n_U}$ in the collection (0) such that $$ x \in B_{x, n_U} \subset U. \tag{1} $$
We try to show that the countable collection $$ \left\{ f \left( B_{x, n} \right) \colon n \in \mathbb{N} \right\} \tag{2} $$ is a local basis at $f(x)$.
Since $f \colon X \rightarrow Y$ is an open map and since each set $B_{x, n}$ in the collection (0) above is an open set in $X$, therefore each image set $f \left( B_{x, n} \right)$ is an open set in $Y$ and hence is also an open set in the subspace $f(X)$ because $f \left( B_{x, n} \right) \subset f(X)$.
And, since each set $B_{x, n}$ contains $x$, therefore each set $f \left( B_{x, n} \right)$ contains the point $y = f(x)$.
Let $V$ be any open set of $f(X)$ such that $y = f(x) \in V$. Then there exists an open set $V^\prime$ of $Y$ such that $$ V = f(X) \cap V^\prime. $$ Then of course $f(x) \in V^\prime$ also, which implies that $x \in f^{-1} \left( V^\prime \right)$.
In fact we even also have $$ \begin{align} f^{-1}(V) &= f^{-1} \left( f(X) \cap V^\prime \right) \\ &= f^{-1} \big( f(X) \big) \cap f^{-1} \left( V^\prime \right) \\ &= X \cap f^{-1} \left( V^\prime \right) \\ &= f^{-1} \left( V^\prime \right). \end{align} $$ Thus we have the equality $$ f^{-1}(V) = f^{-1} \left( V^\prime \right). \tag{3} $$ And from the last line of the preceding paragraph we have $$ x \in f^{-1}(V) = f^{-1} \left( V^\prime \right).$$
Now as $f \colon X \rightarrow Y$ is a continuous map and as $V^\prime$ is an open set of $Y$, so the inverse image $f^{-1} \left( V^\prime \right) = f^{-1}(V)$ is also an open set of $X$. [Please refer to (3) above.]
As $U \colon= f^{-1}(V)$ is an open set of $X$ containing $x$, so there exists a set $B_{x, n_U}$ in the collection (0) above such that the relation (1) above holds, and therefore we can conclude that $$ f(x) \in f \left( B_{x, n_U} \right) \subset f(U) = f\left( f^{-1} (V) \right) \subset V, $$ that is, $$ f(x) \in f \left( B_{x, n_U} \right) \subset V. $$
Thus the collection (2) is a countable local basis at the point $y = f(x) \in f(X)$.
But $y = f(x)$ was an arbitrary point of $f(X)$. Hence $f(X)$ is first-countable.
Am I right?
Case 2. Next, suppose that $X$ is second-countable.
Let $$ \left\{ B_n \colon n \in \mathbb{N} \right\} \tag{4} $$ be a countable basis for the topology of $X$.
Then each set $B_n$ in the collection (4) is an open set of $X$; moreover for each open set $U$ of $X$ and for each point $x \in U$, there exists a set $B_{n_{U, x}}$ in the collection (4) above such that $$ x \in B_{n_{U, x}} \subset U. \tag{5} $$
We seek to show that the countable collection $$ \left\{ f \left( B_n \right) \colon n \in \mathbb{N} \right\} \tag{6} $$ is a basis for the subspace topology of $f(X)$ regarded as a subspace of $Y$.
Let $n \in \mathbb{N}$. Now as $f \colon X \rightarrow Y$ is an open map and as the set $B_n$ in the collection (4) is an open set of $X$, so we can conclude that the image set $f \left( B_n \right)$ is an open set of $Y$ and hence is also an open set of $f(X)$ because $f \left( B_n \right) \subset f(X)$. Thus the collection (6) is a collection of open sets in the subspace $f(X)$ of $Y$.
Let $V$ be any open set of $f(X)$ and let $y$ be any point in $V$. Then there exists an open set $V^\prime$ of $Y$ such that $$ V = f(X) \cap V^\prime. $$ Then of course $y \in f(X)$ and $y \in V^\prime$ which implies that $y = f(x)$ for some point $x \in X$ and that $x \in f^{-1} \left( V^\prime \right)$ also.
As in Case 1 above we have the identity $$ f^{-1}(V) = f^{-1} \left( V^\prime \right). \tag{7} $$ And from the last line of the preceding paragraph we have $$ x \in f^{-1}(V) = f^{-1} \left( V^\prime \right) .$$
Moreover, as $f \colon X \rightarrow Y$ is a continuous map and as $V^\prime$ is an open set of $Y$, so the set $f^{-1}(V) = f^{-1} \left( V^\prime \right)$ is an open set of $X$; let us put $$ U \colon= f^{-1}(V). \tag{8} $$ [Please refer to (7) above.] Then $U$ is an open set of $X$ such that $x \in U$. So there exists a set $B_{n_{U, x}}$ in the collection (4) above such that (5) above holds.
From (5) and (8) above we obtain $$ y = f(x) \in f \left( B_{n_{U, x}} \right) \subset f(U) = f \left( f^{-1}(V) \right) \subset V, $$ and thus $$ y \in f \left( B_{n_{U, x}} \right) \subset V. \tag{9} $$
Thus the countable collection (6) above is a collection of open sets in the subspace $f(X)$ of $Y$ such that for any open set $V$ of $f(X)$ and for any point $y \in V$ there exists a set in the collection (6) for which the relations in (9) above holds. Thus the collection (6) above is indeed a basis for the subspace topology of $f(X)$, by Lemma 13.2 in Munkres.
Therefore the collection (6) is a countable basis for the topology of $f(X)$ regarded as a subspace of $Y$.
Hence $f(X)$ (regarded as a subspace of $Y$) is second-countable.
Am I right?
Are both of my proofs are correct and clear enough presentation-wise? Or, are there problems in either one?
I didn't read the proofs for the simple reason they are way too long.
You can make them way shorter...
For example:
Suppose $X$ is second countable. Define $g: X \to f(X): x \mapsto f(x)$. By construction, $g$ is continuous, open and surjective. Let $\mathcal{B}$ be a countable base for $X$. Then we claim that $g(\mathcal{B}):= \{g(B): B \in \mathcal{B}\}$ is also a countable base.
It is obvious that $|g(\mathcal{B})| \leq |\mathcal{B}|$ so $g(\mathcal{B})$ is countable. Next, let $A$ be an open in $f(X)$. Since $g$ is continuous and $\mathcal{B}$ is a basis for $X$, we can write $g^{-1}(A) = \bigcup \mathcal{B}_0$ for a subcollection $\mathcal{B}_0 \subseteq \mathcal{B}$. Consequently, by surjectivity of $g$
$$A = g(g^{-1}(A)) = \bigcup g(\mathcal{B}_0)$$
and we are done.