Does a continuous function preserve Bolzano-Weierstrass property? 
Let $X,Y$ be arbitrary topological spaces with $X$ satisfying the Bolzano-Weierstrass property (i.e. every infinite subset has some accumulation point) and let $f:X\to Y$ be a continuous function, does $f(X)$ satisfy the B-W property? 

I tried to prove it does by contradiction as follows:
Let's assume $f(X)$ doesn't have the BWP, so there must be some infinite subset $A$ wich has not any accumulation point, hence it's is closed. Because $f$ is continuous, $f^{-1}(A)$ is also closed and infinite and because $X$ has the BWP there is some accumulation point of $f^{-1}(A)$ which is contained in it, let's call $x_0$ one of such points. Now, $f(x_0)$ is an isolated point of A, so $\{f(x_0)\}$ is an open neighborhood of $f(x_0)$ (with the relative topology of $f(X)$ as a subspace of $Y$).
This means that $f^{-1}(f(x_0))$ is an open neighborhood of $x_0$ which contains other points different than $x_0$. But this doesn't lead me to a contradiction since in principle $f$ could be locally constant at $x_0$. Is there some reason why this cannot happen? Or maybe continuous function doesn't actually preserve BWP, if that's the case could someone provide a counterexample? Constant functions doesn't seem to work because they have finite image, so I thought maybe a locally constant function with infinite image, but I couldn't build up an explicit example.
EDIT to avoid future missunderstandings: What I called 'accumulation point' above is what usually is refered in english as 'limit point'. Explicitly $a$ is an accumulation (limit) point of a set $A$ of a topological space $X$ if every open set containing $a$ intersected with $A$ contains some point different than $a$
 A: This is actually false if "accumulation point" = "limit point" where $x$ is a limit point of $A$ iff every neighbourhood of $x$ intersects $A\setminus\{x\}$.
Your B-W property is then called "limit point compact"
A counterexample: Let $Y = \mathbb{N}$ as a discrete space and let $X = Y \times \{0,1\}$ where $\{0,1\}$ has the indiscrete (trivial) topology with no non-trivial open sets. Let $f= \pi_Y$ be the projection onto $Y$, which is continuous. 
$X$ is limit point compact: if $A \subseteq X$, and $(x,i) \in A$, then $(x,i')$ is a limit point of $A$, where $i' \in \{0,1\}\setminus\{i\}$, because any product open set that contains $(x,i)$ also contains $(x,i')$ and vice versa. But $Y$ is infinite discrete so not limit point compact.
If $X$ is $T_1$ then (here our example is not even $T_0$) limit point compact is equivalent to "strong limit point compact": every infinite set $A$ has an ($\omega$-)accumulation point, i.e. an $x \in X$ such that every neighbourhood of $x$ contains infinitely many points from $A$. This property is preserved by continuous functions:
Suppose $X$ is strong limit point compact, and $f:X \to Y$ is continuous.
Let $A \subseteq f[X]$ be infinite and let $\{y_n: n \in \omega\}$ be an enumeration ($y_n \neq y_m$ for $n \neq m$) of a countably infinite subset of $A$ which means that we have for all $n$ we have $x_n \in X$ such that $f(x_n) = y_n$. Then clearly, $n \neq m$ implies $x_n \neq x_m$ as well, so $\{x_n: n \in \omega\}$ is an infinite subset of $X$. So this has a strong limit point $x$. Now let $y = f(x) \in f[X]$.
If $O$ is any open set containing $y$ then $f^{-1}[O]$ is open by continuity and contains $x$, so there are infinitely many $n$ with $x_n \in f^{-1}[O]$. All those $n$ also obey $y_n = f(x_n) \in O$. So $y$ is a strong limit point of $A$, and $f[X]$ is strong limit point compact.
So your original statement is true for $T_1$ spaces $X$. We then get even a (slightly) stronger property than B-W in $Y$ (without assuming anything extra on $Y$), namely strong limit compact.
A: Let $f(x_n)=y_n\in f(X)$ be a sequence of different points. Then, $x_n$ are different too. Passing to a subsequence, there is an $x$ such that $x_n$ accumulates in $x$. Let $V\ni y=f(x)$ be an open subset of $Y$. Then $U=f^{-1}(V)$ is open in $X$ and contains $x$. Therefore some $x_n$ lies in $U$. Therefore $y_n=f(x_n)$ lies in $V$. This means that $y_n$ accumulates in $y$.  
A: Continuity alone has nothing to do with Bolzano-Weierstrass.
Returning to definitions often helps. You need to show that the range of $f$ is well-ordered, and, in addition, continuous.
