So I have to show both directions. For the if direction, I began as below.
Suppose such a $c$ exists for each pair of $x \in X$ and its corresponding $y = Lx \in Y$. I have to show that the image of an open set containing $x$ by $L$ is an open set in $Y$. Let $x \in U$ where $U$ is an open set in $X$. Also, since $Y$ is subjective, there is a $y \in Y$ such that $y=Lx$. I have to find an open ball in $Y$ containing $y$. Such a ball would be of the form $\{y' \in Y, \|y'-y\|<r \}$ for some $r>0$. Plus, again due to subjectivity, every $y'=Lx'$ for some $x'\in X$. Thus, $$r>\|y'-y\|=\|Lx'-Lx\| = \|L(x'-x)\| > \frac{\|x-x'\|}{c}.$$ Which gives me $\|x-x'\|< rc$. Am I on the right track for this part? How can I conclude from this that $L$ maps each open ball of radius $rc$ in $X$ into another open ball of radius $r$ in $Y$? And is it so?
As for the other direction, I guess that I should somehow use the fact that the image of the open unit ball under any open mapping contains another open ball. That is $B_Y(0,\rho) \subset L\Big( B_X(0,1) \Big)$ for some $\rho >0$.
I really appreciate any help on both sides (the opposite direction seems more challenging!) Thanks. Edit: $X$ and $Y$ are not necessarily Banach spaces.
For the other direction, if $L$ is an open mapping, then we can find $ \delta>0 $ such that $ L(B_X) \supset \delta B_Y $, where $B_X$ and $B_Y$ are the open unit balls of $X$ and $Y$ resp.
Let $ y \in Y $. If $y=0$, we have nothing to prove. Therefore, we may assume that $ y \neq 0 $. Note that $ \frac{\delta }{2} \frac{y}{\ ||{y}||} \in L(B_x) $ and so there exists $ z \in B_X $ such that $ \frac{\delta }{2} \frac{y}{||{y}||}=L(z)$. In other words, $ y=L(\frac{\delta }{2} ||y|| z ) $. If we let $ x =\frac{\delta }{2} ||y|| z $, then $ ||x|| \leq \frac{\delta}{2} ||y||$. Take $c=\delta/2$.
Edit: For the original direction, notice that it suffices to show that there exists $\delta >0$ such that $ L(Β_X) \supset \delta B_Y $ (which follows immediately from the hypothesis). Indeed, suppose such a $\delta$ exists. Let $U \subset X$ be open and let $y_0 \in L(U)$. Then, $y_0 = Lx_0 $ for some $x_0 \in U$. Since $U$ is open, there exists $r>0$ such that $B_X(x_0,r) \subset U$, or equivalently, $x_0 +B_Χ(0,r) \subset U$. Hence, $$ y_0 +L(B_X(0,r)) \subset L(U).$$ Since $ L(B_X(0,r)) \supset \delta Β_Y (0,r) $ we have that $$ B_Y(y_0,\delta r) \subset L(U),$$ that is, $L(U)$ is open in $Y$.