Assuming that you are using the standard metric, injectivity can be shown as follows: Take $a,b\in\mathbb R,a<b$, such that $f(a)=f(b)$. Then $f$ attains its minimum $m$ and maximum $M$ in $[a,b]$, at $x_m,x_M\in[a,b]$, respectively. If $x_m,x_M\in\{a,b\}$, then $f$ is constant on $[a,b]$ and hence not open, contradiction. Thus, $x_m\in(a,b)$ or $x_M\in(a,b)$, and we assume the former. Then $f((a,b))=[m,M]$ (if $x_M\in(a,b))$ or $f((a,b))=[m,M)$ (if $x_M\in\{a,b\})$, but in any case, $f((a,b))$ is not open, again a contradiction. Thus, $f$ is injective and therefore strictly monotone. Now it is clear that $f$ is a homeomorphism.
For the second question consider $f:S^1\to S^1,\ z\mapsto z^2$. Then $f$ is clearly surjective. It is open, since if we take $I:=\{e^{it}\ |\ t\in(a,b)\}$, then $f(I)=\{e^{2it}\ |\ t\in(a,b)\}$. But it is not injective, since $f(1)=1=f(-1)$.
Best Answer
Any surjection that attains a local extremum suffices. Consider for instance $$ f(x) = x(x-1)(x-2) = x^3 - 3x^2 + 2x $$ Note that $f$ is not an open map, since the interval $(0,1)$ is mapped to an interval of the form $(0,a]$, which is not open.