Do continuous functions "preserve limits" between spaces where uniqueness of limits isn't guaranteed? As far as I know (I think I am not mistaken), the following is true: "Given two Hausdorff spaces $X$ and $Y$ and $f$ a mapping between them, the following two statements are equivalent:

*

*$f$ is continuous

*Given a sequence $(x_n)$ converging to $x$, then the sequence $(f(x_n))$ converges to $f(x)$"

My question is: can this be generalized in the case of "convergence without uniqueness"? The proposed generalization being:
"Given two topological spaces $X$ and $Y$, a mapping $f$ between them is continuous if and only if, for any sequence $(x_n)$ converging to a set of points $A$, the corresponding sequence $(f(x_n))$ is convergent, and converges to the set $f(A)$"
I specially care about the "only if" part, because I need it for further results that I'm trying to achieve. Of course, I already come with a proof for this part that I think is correct, but I ask here this question to put this proof to the test of collective scrutiny. The proof in question:
"Let $X$ and $Y$ be topological spaces, let $f$ be a continuous mapping between them. Let $(x_n)$ be a sequence converging to a set $A$, let $x$ be a point in A. Then for each open $G$ containing x, there exists a certain $n_0$ such that for any $n > n_0$, $x_n$ is in $G$.
Let $H$ be an open in $Y$ such that $f(x)$ is in $H$. Then $f^{-1}(H)$ is an open in $X$ containing $x$. Then there exists $n_0$ with the above property for $f^{-1}(H)$. But
$$x_n\in f^{-1}(H) \Leftrightarrow f(x_n) \in H$$
Then $f(x)$ is a limit point of the sequence $(f(x_n))$."
Is this proof correct? Also, is the other implication true? Thank you for your attention.
 A: The converse statement is false. Let $\tau_{\rm{usual}}$ be the usual topology on $\mathbb R$ and $\tau_{cc}$ be the co-countable topology on $\mathbb R$, that is :
$$\tau_{cc} = \{U\subset \mathbb R~|~U= \emptyset \text{ or } X\backslash U\text{ is at most countable} \}$$
Then, $\tau_{cc}$ is not Hausdorff and a sequence in $(\mathbb R , \tau)$ is convergent if, and only if, it is eventually constant.
Therefore, the identity $\text{id}:(\mathbb R,\tau_{cc}) \to (\mathbb R, \tau_{\rm{usual}})$ sends convergent sequence to convergent sequences (and limit points to limit points) but is not continuous.
A: The statement under 2 just says that if $x$ happens to be one of the points that a sequence $(x_n)_n$ converges to, then $f(x)$ is also one of the points that $(f(x_n))_n$ converges to.
Or in logical notation:
$$\forall (x_n)_n \subseteq X: \forall x \in X: (x_n \to x) \implies (f(x_n) \to f(x))\tag{1}$$
There is no assumption of unicity of limits at all. Moreover, having $(1)$ it is certainly true that if $A$ is the set of all limits of $((x_n))_n$ (which is, I think, what you mean by "$(x_n)$ converges to $X$", though this is not a usual definition) then $f[A]$ is a subset of all limits of $(f(x_n))_n$; this follows straight from $(1)$, so there is no "generalisation" to be had here.
Your proof for $(1)$ (i.e. statement 2) from the continuity is fine. Hausdorffness on $X$ or $Y$ is not needed at all.
As already said by @SolubleFish, the reverse, i.e. 1 from 2 is not in general true, though it is true when $X$ is a sequential space; one of the reasons this notion was introduced. So e.g. in the realm of metric spaces $X$, $2 \to 1$ does hold, while $1 \to 2$ holds for all spaces.
