How to show $f(x)=x^{2}$ is a closed map? The actual question was $\\$
"Show $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $f(x)=x^{2}$ is closed but not open."
For the latter one, $f((-1,1))=[0,1)$ so $f$ is not an open map, but I have trouble showing that it is closed. 
 A: Here is a more general statement that will perhaps interest you.
Actually, each polynomial $P$ on $\mathbb{R}$ is a closed function.
Let $F$ be a closed subset of $\mathbb{R}$. We will show that $P(F)$ is closed. Take $y_n$ a convergent sequence of $P(F)$ and $y$ its limit. We have to show that $y \in P(F)$. By definition, there is $x_n$ in $F$ such that $y_n = P(x_n)$. Since $y_n$ is bounded (since it is convergent), $x_n$ is also bounded because a polynomial goes to infinity at $\pm \infty$. Because $x_n$ is a bounded sequence of $\mathbb{R}$, one can extract a subsequence $x_{k(n)}$ of $x_n$ which converges. Let us note $x$ its limit. Since $F$ is closed, one has $x \in F$. Since $P$ is continuous (it is a polynomial), we have $$y_{k(n)}=P(x_{k(n)}) \to P(x).$$ However $y_{k(n)}$ does also converge to $y$. Hence by unicity of the limit, we have $$y = P(x) \in P(F)$$ and the conclusion follows.
A: Let $I=[a,b] \subset \Bbb R$ a closed interval, where $a \le b$.
There are 3 cases: (1) $a\le b \le 0$, (2) $a \le 0 \le b$, (3) $0 \le a \le b$.
In case (1), $f(I) = [b^2, a^2]$. 
In case (2), $f(I) = [0, \max\{a^2,b^2\}]$.
In case (3), $f(I) = [a^2, b^2]$.
In any case, $f(I)$ is closed and therefore $f$ maps any closed interval to a closed interval, hence it is closed.
