The problem is this.
$prove\ f(\bigcap_\alpha E_\alpha)=\bigcap_\alpha f(E_\alpha)$
So I tried it below.
$(1)\ f(\bigcap_\alpha E_\alpha)\subset \bigcap_\alpha f(E_\alpha)$
Suppose $y\in f(\bigcap_\alpha E_\alpha)$
$\Rightarrow \exists x\in \bigcap_\alpha E_\alpha\ s.t. y=f(x) \ for \ all\ a$
$\Rightarrow \exists x\in E_\alpha\ s.t. y=f(x)\ for\ all\ a$
$\Rightarrow \exists x\in E_\alpha\ s.t. y\in f(x)\ for\ all\ a$
$\Rightarrow y\in f(E_\alpha)\ for\ all\ a$
$\Rightarrow y \in \bigcap_\alpha f(E_\alpha)$
$(2)\ \bigcap_\alpha f(E_\alpha)\subset f(\bigcap_\alpha E_\alpha)$
Suppose $y \in \bigcap_\alpha f(E_\alpha) \Rightarrow y \in f(E_\alpha)\ for\ all\ a$
$\Rightarrow \exists x\in E_\alpha\ s.t. y=f(x) \ for \ all\ a$
$\Rightarrow \exists x\in E_\alpha\ s.t. y \in f(x)\ for\ all\ a$
$\Rightarrow \exists x\in \bigcap_\alpha E_\alpha\ s.t. y\in f(x)\ for\ all\ a$
$\Rightarrow y\in f(\bigcap_\alpha E_\alpha)$
Is there any error?
Your proof does not work, as we cannot conclude $y\in f(x)$ from $y=f(x)$.
Here is a correct proof: Let $x\in \bigcap_\alpha E_\alpha$, then $x\in E_\alpha$ for all $\alpha$. Therefore, $f(x)\in f(E_\alpha)$ for all $\alpha$, and we have $f(x)\in \bigcap_\alpha f(E_\alpha)$. By the definition of image under $f$, we have $f(\bigcap_\alpha E_\alpha)\subseteq\bigcap_\alpha f(E_\alpha)$.
The reverse inclusion does not hold in general (A counterexample by copper.hat works.) If $f$ is one-to-one, however, we can show the reverse inclusion:
Let $f$ be one-to-one and $y\in \bigcap_\alpha f(E_\alpha)$. For $\alpha\neq \beta$, we have $x\in E_\alpha$ and $z\in E_\beta$ such that $f(x)=y=f(z)$. Then $x=z$ as $f$ is one-to-one, so $x\in E_\beta$. Since $\beta$ is arbitrary, we can conclude that $x\in\bigcap_\alpha E_\alpha$ and $y=f(x)\in f(\bigcap_\alpha E_\alpha).$