Let $f:\Omega\to\mathbb{R}$ be a continuous function over convex domain $\Omega$. Is the set $$ f\left(\Omega\right)=\left\{ y:y=f\left(x\right):x\in\Omega\right\} $$ Is the set $f(\Omega)$ convex ?
Assume $$ \begin{cases} y_{1}=f\left(x_{1}\right)\\ y_{2}=f\left(x_{2}\right) \end{cases} $$ Looking at points between $y_1,y_2$ we get $ y_{t}=y_{1}+\left(y_{2}-y_{1}\right)t $ where $0\le t\le 1 $ $$ y_{t}=f\left(x_{1}\right)+\left[f\left(x_{2}\right)-f\left(x_{1}\right)\right]t=\left(1-t\right)f\left(x_{1}\right)+tf\left(x_{2}\right)$$
The continuous property says $\forall \varepsilon>0$ if there exists $\delta$ such that $$ \left|z-x_{1}\right|<\delta $$ then $$\left|f\left(z\right)-f\left(x_{1}\right)\right|\le\varepsilon $$ I tried setting $ z=x_{1}+\alpha\left(x_{2}-x_{1}\right) $, but got nowhere.
if $f$ is invertible then we can easily find an $x $ such that $f(x)=y_t$
Is there a way to prove exists such an $x$ using the continuous property ?
Your set is written incorrectly, with two $:$ symbols. I would write it as $$f(\Omega) = \{ y \in \mathbb{R} : \exists x \in \Omega \text{ satisfying } f(x) = y \}.$$
Now for the proof:
Without loss of generality, we may assume $$y_1 < y_t < y_2.$$ Define the function $$g(c) = f((1-c)x_1 + cx_2),$$ for $t \in [0,1]$ and note that $g(0) =y_1$ and $g(1) = y_2$. Note also that $g$ is well defined, since $\Omega$ is convex, and that $g$ is continuous. By the intermediate value theorem, there exists some $c^* \in [0,1]$ such that $g(c^*) = y_t$. Hence, $y_t \in f(\Omega)$.