Proof verification: Image of the intersection is the intersection of the images $$ f\left(\bigcap\limits_{\lambda \in \wedge} A_{\lambda}\right) = \bigcap\limits_{\lambda \in \wedge} f(A_\lambda)$$
$(\rightarrow)$
Let $b \in f(\bigcap\limits_{\lambda \in \wedge} A_{\lambda})$
$\rightarrow b=f(a)$ for some $a \in (\bigcap\limits_{\lambda \in \wedge} A_{\lambda})$.
Since $a \in (\bigcap\limits_{\lambda \in \wedge} A_{\lambda}) \rightarrow a \in A_{\lambda}$ for some $\lambda \in \wedge$.
Since $b=f(a)$, then $b \in f(A_{\lambda})$ for some $\lambda \in \wedge$.
$\rightarrow b \in \bigcap\limits_{\lambda \in \wedge} f(A_\lambda)$
$\rightarrow f(\bigcap\limits_{\lambda \in \wedge} A_{\lambda}) \subset \bigcap\limits_{\lambda \in \wedge} f(A_\lambda)$.
$(\leftarrow)$
Let $b \in \bigcap\limits_{\lambda \in \wedge} f(A_\lambda) \rightarrow b \in f(A_\lambda)$ for some $\lambda \in \wedge$ 
$\rightarrow (*) b=f(a)$ for every $a \in A_\lambda$ and $\lambda \in \wedge$ $\rightarrow b=f(a)$ for every $a \in \bigcap\limits_{\lambda \in \wedge} A_\lambda$
In order for $(*)$ to be true, $f$ must be $1-1$
$b=f(a)$ for every $a \in \bigcap\limits_{\lambda \in \wedge} A_\lambda \rightarrow b \in f(\in \bigcap\limits_{\lambda \in \wedge} A_\lambda)$
Both $(\rightarrow)$ and $(\leftarrow)$ imply that $$ f\left(\bigcap\limits_{\lambda \in \wedge} A_{\lambda}\right) = \bigcap\limits_{\lambda \in \wedge} f(A_\lambda)$$
This problem is in Willards, "General Topology." Please let me know if this solution is accurate.
 A: [Added later:]
You miss the point of the original exercise: 



One should make a correct statement first.
The statement in your question is incorrect. (To make it correct, either replace the equal sign with $\subset$ or replace $f$ with $f^{-1}$. Or you need extra assumptions on $f$.) See this question: 
Does an arbitrary function preserve arbitrary intersections?

$$ f\left(\bigcap\limits_{\lambda \in \wedge} A_{\lambda}\right) = \bigcap\limits_{\lambda \in \wedge} f(A_\lambda)$$
$(\rightarrow)$ (OK. You mean $\subset$.)
Let $b \in f(\bigcap\limits_{\lambda \in \wedge} A_{\lambda})$
  $\rightarrow b=f(a)$ for some $a \in (\bigcap\limits_{\lambda \in \wedge} A_{\lambda})$. (You could use proper English words to make your argument more readable. )
Since $a \in (\bigcap\limits_{\lambda \in \wedge} A_{\lambda}) \rightarrow a \in A_{\lambda}$ for some $\lambda \in \wedge$. (This is a mistake. "for some" should be "for every". Your argument breaks down. I would stop reading from here.)
$(\leftarrow)$ (This direction is wrong.)


This problem is in Willards, "General Topology." Please let me know if this solution is accurate. (Not yet.)
A: Let $b \in f(\bigcap\limits_{\lambda \in \Lambda}A_{\lambda})$, then $b = f(a)$ for some $a \in \bigcap\limits_{\lambda \in \Lambda}A_{\lambda}$, i.e. $\forall \lambda \in \Lambda$ $a \in A_{\lambda}$, hence $\forall \lambda \in \Lambda$ $f(a) \in f(A_{\lambda})$. So, $b \in \bigcap\limits_{\lambda \in \Lambda}f(A_{\lambda})$. This gives $f(\bigcap\limits_{\lambda \in \Lambda}A_{\lambda}) \subseteq \bigcap\limits_{\lambda \in \Lambda}f(A_{\lambda})$.
But the second direction does not work. Imagine $f$ to be constant function, but $A_{\lambda}$ be a partitioning of the domain of $f$. So image of intersection is empty, while intersection of images is not.
