Let the function $f \colon \mathbb{R} \longrightarrow \mathbb{R}$ be defined by $f(x) := x^2$ for all $x \in \mathbb{R}$, where the set $\mathbb{R}$ of real numbers has the usual topology. Then $f$ is a closed map. [Am I right?]
Let $A$ be any closed set in (the domain space) $\mathbb{R}$. We need to show that $f(A)$ is also a closed set in (the co-domain space) $\mathbb{R}$. Let $y$ be any point in the closure of $f(A)$. Then there exists a sequence $\left( y_n \right)_{n \in \mathbb{N}}$ of points of $f(A)$ converging to this $y$.
As $f(A) = \left\{ x^2 \colon x \in A \right\}$, so we must have $f(A) \subseteq [0, +\infty)$. So if $y$ were less than $0$, then there would be no elements of $f(A)$ in the open set $(-\infty, 0)$ containing $y$, and thus there would be no sequence of points of $f(A)$ converging to $y$. So we must have $y \geq 0$. [Am I right?]
For each $n \in \mathbb{N}$, as $y_n \in f(A)$, so we must have $y_n = f \left( x_n \right) = x_n^2$ for some element $x_n \in A$. Thus we can conclude that the sequence $\left( x_n^2 \right)_{n \in \mathbb{N}}$ converges to $y$, that is, $$ \lim_{n \to \infty} x_n^2 = y, $$ and since $y \geq 0$, we must also have $$ \lim_{n \to \infty} x_n^2 = \left( \sqrt{y} \right)^2. $$ So there exists a subsequence $\left( x_{n_k} \right)$ of $\left( x_n \right)$ which converges either to $\sqrt{y}$ or to $-\sqrt{y}$. Thus we have either $\sqrt{y} \in \overline{A}$ or $-\sqrt{y} \in \overline{A}$. But since $A$ is a closed set (in $\mathbb{R}$), we can conclude that either $\sqrt{y} \in A$ or $-\sqrt{y} \in A$, and hence in either case we can conclude that $$ y = \left(\pm \sqrt{y} \right)^2 \in f(A), $$ from which it follows that $f(A)$ is a closed set in (the co-domain space) $\mathbb{R}$. [Am I right?]
Is each and every step in the above argument correct and clear enough? Or, are there issues?
Last but not least, how to show that $f(A)$ is closed by showing that $\mathbb{R} \setminus f(A)$ is open in $\mathbb{R}$?