How easily to check: $f$ is a closed map? Assume that $f:\mathbb R\to\mathbb R$ is a function where $\mathbb R$ is the real number and the usual topology is defined on $\mathbb R$.
I have two questions:
   1. Let $C$ be closed. Then what is the proper form of $C$? For example, without the loss of generality, open set is replaced by the open interval.
   2. To check that $f$ is a closed map, we need some information: $f(C)$ is closed. How to easily check this? 
 A: As I said in the comments, it will depend on the map $f$, in general.
In the case of $f(x) = x^2$ (also see comments) we can write $\mathbb{R}$ as $A \cup B$, where $A = \{x: x \le 0 \}, B = \{ x: x \ge 0\}$, so $A$ and $B$ are closed and $f | A$ and $f|B$ are homeomorphisms onto $[0, \infty)$. If then $C$ is closed, then $f[A \cap C]$ is closed in $[0,\infty)$, as homeomorphisms are closed maps and $A \cap C$ is closed in $A$, and likewise for $f[B \cap C]$. Hence their union which equals $f[C]$ is closed in $[0, \infty)$ as well and thus closed in $\mathbb{R}$ (closed in a closed sets is closed in the whole space). So $f$ is a closed map.
A: There are several ways to check that a map is closed.
The formula $\overline{f(A)}\subseteq f(\overline A)$ is a description of closed maps. However, since the proof that this formula is equivalent to closedness of $f$ is very easy, I think you could as well show directly that your $f$ is closed.
Here are some hints:


*

*A characterization of closedness of $f$ is: For each $y\in Y$ and every open $U\supseteq f^{-1}(y)$ there is an open $V\ni y$ with $f^{-1}(V)\subseteq U$.

*Closedness is local in the codomain. This means that for each point $y\in Y$ there is a neighborhood $V$ such that the restricted $f:f^{-1}(V)\to V$ is closed. Equivalently, $Y$ is covered by the interiors of subsets $(V_i)_{i\in I}$, where $f_i:f^{-1}(V_i)\to V_i$ is closed for each $i\in I$. As the restriction $f:S\to f(S)$ of a closed map to a saturated set $S$ is again closed, this condition is also necessary.
The second point can be applied if $f$ is a proper map (preimages of compact sets are compact) and the codomain is locally compact Hausdorff or, more generally, each point in $Y$ has a compact Hausdorff neighborhood. In that case it is easy to prove that $f$ is a closed map. For example, a continuous map from $\Bbb R$ to $\Bbb R$ is closed if the preimage of a bounded set is bounded, so $f(x)=x^2$ is closed.
Yet another fact: If $f:X\to Y$ is proper and $Y$ is Fréchet-Urysohn and UCC, then $f$ is closed. UCC (unique convergent clustering) means that every convergent sequence has a unique cluster point.
A: To check that $f(C)$ is closed you can either prove that all the limit points of $f(C)$ lie in $f(C)$ or $\mathbb{R}\setminus f(C)$ is open. The second one is easy most of the times but it really depends on what is $f$.
Also open sets are unions of open intervals not only open intervals.
