Can someone check my proof please for Rudin chapter 4 I want to prove:

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that $$f(\overline{E}) \subset \overline{f(E)} $$
  for every set $E\subset X$. ($\overline{E}$ denotes the closure of $E$.)

If $x$ is in  $\overline{E}$, then its either a limit point of $E$ or a point of $E$ or both.  If its a point of $E$, then $f(x)$  is contained in $f(E)$. and $$f({E}) \subset \overline{f(E)} $$. If it is a limit point, there are infinitely many elements surrounding it. This means that since $f$ is continuous, $|f(x_1)-f(x)|< $$\epsilon$ and we can make $\epsilon$ very close to zero and there will be always elements $x_1$ in E to satisfy $|x_1-x|<$$\delta$. So either $f$ is a constant function to make $|f(x_1)-f(x)|=0< $$\epsilon$ or the $f(x_1)$ have to approach $f(x)$ as they the $x_1$ get closer and closer to $x$. This means that $f(x)$ is a limit point of $f(E)$ so is contained in $\overline{f(E)} $ 
Thank you for your time
 A: 
If its a point of $E$
  , it is contained in$
f
(
E
)$
  . 

Note $E\subset X$ and $f(E)\subset Y$. Are you claiming $x$ is in $X\cap Y$?
The rest of your proof has traces of the right idea, but its quite hard to read. I would suggest that instead of "and we can make $\epsilon$ very close to zero", "there will always be elements...", you replace this with the more standard phrasing 

By the continuity of $f$ at $x\in X$, for every $\epsilon>0$, there exists $\delta>0$ and $x_1\in X$ such that...

Also, while its true that $f$ is either a constant function or not a constant function, this splitting of cases is not helpful at all. First, the behavior you seem to be worried about can still happen if $f$ is not constant (e.g. constant only on a small neighbourhood of $x$.) Second, I suspect the reason you care to make the split is because you chose to write it out in english (" get closer and closer"). Again, you can use more standard phrasing, like 

we see that  (e.g. by choosing $\epsilon = 1/n$ in the above for each $n\in\mathbb N$) there exists a sequence of points $x_n\in X$ such that $x_n\to x$ and $f(x_n)\to f(x)$.

Observe that the case where $f(x_n) $ are all (eventually) equal to $f(x)$ is already covered by this. It is also easier (to me, at least) to verify that $f(x)$ is a limit point from this.
