Let $A=f(B)\subset X$ where $B=\{{(x,y) \in \mathbb{R}^2 \mid 1 \leq x^2+y^2 \leq 2 \}} $, $X$ is arbitrary topological space and $f: \mathbb{R}^2 \to X$ is an arbitrary continuous map. Then $A$ is compact and connected but not closed and open.
I have a doubt that if $A$ is compact then it is closed, but the answer says only compact and connected. Please help me as I have learnt topology several years back and I am not that strong in it now.
On
Compactness and connectedness are preserved by continuous maps, thus the compactness and connectedness of $B$ imply the same properties on $A$, its image.
It is not necessarily open as the identity $\mathbb{R}^2 \to \mathbb{R}^2$ shows.
In most common cases, to be honest, the image is indeed closed. The hypothesis you have to require is $X$ to be an hausdorff space: every two points can be separated by opens. Precisely, for every $x,y \in X$ there exist disjoint opens $U,V$ such that $x \in U, y \in V$. This property is shared by almost everything geometric that come to your mind.
Still, there are counterexamples to "compact $\Rightarrow $ closed" without the hausdorff property. Define $X$ in this way. Its points are real numbers plus a special point, $z$. Its closed sets are a finite number of real points ( $z$ is not allowed) or the entire $X$. Then the one-point-set $\{z\}$ is of course compact because every open covering is made of one open (it is just one point!), but it is not closed: its closure is the whole $X$! The other closed sets do not contain $z$.
This example arise often in algebraic geometry. It is like we have added the real line itself as $z$ to the set of real numbers, and we declared it near to every point (every open contains $z$). This allow algebraic geometrists to make interesting tricks :)
In hausdorff case, the proof that a compact is closed is easy. Try to figure it out geometrically.
Let $K \subset X$ be a compact set, and let $p\not \in K$. We must show that there is an open set that contains $p$ and does not intersect $K$ (i.e., the complement is open). By hp, for every $x \in K$ there exist $x \in U_x , p \in V_x$ open sets which separate $x,p$. $\{U_x \}_x$ is an open covering of $K$, thus we can extract a finite covering $U_1, .., U_n$. Then the corresponding $V_i$ have intersection $V= \bigcap V_i$ which contains $p$ and it is disjoint by $\bigcup U_i \supset K$. Yuppi!
$A$ can be open and closed: take $X=\{0\}$ with any topology you like. Then we have that $f$ is constant, so continuous, and $A=f(B)=\{0\}=X$ is open and closed as it is the space itself.
Now assuming one of the equivalent statements: $f$ is injective or $f^{-1}(A)=B$:
Suppose that $A$ is closed and open, then due to the continuity of $f$ we would have that $B=f^{-1}(A)$ is closed and open, but obviously $B$ is not open is $\mathbb{R}^2$ (take for example any open ball around $(2,0)$, this will always have a non-empty intersection with $B^c$), so we have a contradiction.
Remarks:
OP probably already knew it, but the image of a compact set under continuous function $f$ is also compact. Exactly the same holds if you replace the words compact by the words connected.
Taken from one of my comments: Here is the proof of compact implies closed in Hausdorff spaces and here is a counterexample for general topological spaces.
Special thanks to @AndreasBlass for noting that injectivity is needed to make my proof hold.