Show that if $x$ is a limit point of $A\subset X$ and $f:X \rightarrow Y$ is continuous then $f(x)$ is a limit point of $f(A)$ Suppose $f:X\rightarrow Y$ is a continuous function between the topological spaces $X$ and $Y$. Suppose that: $A \subset X$ and that  $x$ is a limit point of $A$.

Show that $f(x)$ is a limit point of $f(A)$

By a limit point $c$ of $B$ I mean that every neighborhood of $c$ intersects $B$ in a point $d$ such that $c\not = d$

What I have done is that: I supposed that $x$ is a limit point of $A$ but $x\not\in A$ and showed the required. but in the case when $x \in A$. My trial goes as follows:

Suppose $W$ was a neighborhood of $f(x)$ so $f^{-1}(W) \text{is a neighborhood of } x$. So $ f^{-1}(W) \cap A \not=\emptyset$ because $x$ is a limit point of $A$. Say $y\in f^{-1}(W) \cap A$. So $f(y)\in A$ and $f(y)\in W$. Now, if  $x\not\in A$ then $f(x) \not\in f(A)$ but $f(y)\in A$ so $f(y)\not = f(x)$. So $W$ intersects $f(A)$ in a point $f(y)$ which is different of $f(x)$

If $x\in A$, it may happen that $f$ is not injective and happen that  $f(x)=f(y)$ in which case I can't show that f(x) is a limit point of $f(A)$.

I've tried to came with a counter-example using constant functions on real functions. but it didn't works

What to do? any hints?
As the question was answered in comments and there is no one who added a correct answer to close the question. I'm adding this answer (I got the idea of the  answer from This comment)

The statement is false, here is a concrete example:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ where, $f(x)=2$ for all $x\in \mathbb{R}$. So $f$ is a constant function.

Note that we denote the set of limit points of $H$ by $H'$.

Let $A = [0,1]$.
So, $A' = [0,1]$.

$f(A)=f([0,1])=\{2\}$, but $f(A)'=\emptyset$ because {$2$} has no limit points as if $x\not = 2$  is a real number then there is always an interval $C$ containing $x$ but not contaning $2$.

So Although $1$ is a limit point of $A$, $f(1)=2$ is not a limit point of $f(A)=\{2\}$.

As the question was answered in comments and there is no one who added a correct answer to close the question. I'm adding this answer (I got the idea of the  answer from comments).
$f(x)$ is a limit point of $f(A)$ if and only if every open neighborhood $V\subseteq Y$ of $x$ intersects $f(A)$.

We show the contrapositive. Assume that $f(x)$ is not a limit point of $f(A)$. Then there is an open neighborhood $V_0$ of $f(x)$ such that $V_0\cap f(A) = \emptyset$. Since $f:X\to Y$ is continuous, the preimage $f^{-1}(V_0)$ is an open neighborhood of $x$. 

Claim:  $A \cap f^{-1}(V_0) = \emptyset$. Proof of claim: Suppose $z \in A \cap f^{-1}(V_0)$, then $f(z) \in f(A) \cap V_0$ contrary to assumption.

Now the claim implies that $x$ is not a limit point of $A$ and we're done.