Why is it sufficient to show that $f:X\rightarrow Y$ is open iff every image of an openset contains some nonempty openset I can't wrap my head around the concept that to prove that $f:X\rightarrow Y$ is open, it is sufficient to show that every image of an openset $U\subset X$, $f(U)$ contains some nonempty openset $V\subset Y$. How can this be implied from the definition of an open mapping. That is,  $f:X\rightarrow Y$ is open iff for every openset $U\subset X$, $f(U)$ is open.
 A: You are missing a crucial point! You need to quantify over the elements of $U$; that is $f$ is open if for all open $U$ and all $x\in U$ there exists some open neighborhood $V$ of $f(x)$ contained in $f(U)$.
Recall that a subset $U$ is open iff for all $x\in U$ there exists some open neighborhood $V_x\subseteq U$ of $x$. Indeed, in such a case
$$
U = \bigcup_{x\in U}V_x,
$$
and since every $V_x$ is open, so is $U$. Now, apply this knowledge to $f(U)$.
A: Your statement is false as stated - it can't "sense" what happens at points, which allows for functions like
$$f(x)=x^2$$
where the image of every open set contains an open set, but might not be open.
You can salvage this statement by requiring that this condition somehow apply to all points equally:

A function $f$ is open if, for every $x$ and every open set $U$ containing $x$, there is some open set $V$ contained inside of $U$ which contains $f(x)$.

This is straightforwards to prove - this condition means that every point in $f(U)$ has an open neighborhood in $f(U)$, but then $f(U)$ has to be the union of these neighborhoods - hence must open.
