show that $f$ is continuous at $x$ if: $f(x) ∈ V$ and $V ∈ T_Y ⇒ x \notin \overline {(f ^{−1}(V ))^c}$ Let $(X, T_X)$ and $(Y, T_Y )$ be topological spaces and let $f : X → Y$ be
a function. Let $x ∈ X$. 
I want to show that $f$ is continuous at $x$ if:
$f(x) ∈ V$ and $V ∈ T_Y ⇒ x \notin \overline {(f ^{−1}(V ))^c}$
. 
Then using this result (or otherwise), to show that $f$ is continuous (on $X$)
if and only if $f(\overline{A}) \subseteq \overline{f(A)}$ for any subset $A$ of $X$.
Hmm so does $x \notin \overline {(f ^{−1}(V ))^c}$ imply $x \in \overline{X \setminus f^{-1}[V]}$ which means $V$ is a neighbourhood of $f(x)$ such that $f^{-1}[V]$ is a neighbourhood of $x$?  And for the second part, referring to $f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \bar A$, $f(a)\in \overline{ f(A)}$., can I say assuming $f$ is continuous. Given a subset $A$, we have $f(A)\subseteq \overline{f(A)}$, which means $A\subseteq f^{-1}f(A)\subseteq f^{-1}\overline{f(A)}$. This last set is closed, so $\bar A\subseteq  f^{-1}\overline{f(A)}$, which means $f(\bar A)\subseteq \overline{f(A)}$.
Then conversely, assuming that $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$ , let $F$ be closed in $Y$. Then $f(\overline{f^{-1}(F)})\subseteq \overline{f({f^{-1}(F)})}\subseteq \bar F=F $. Thus $\overline{f^{-1}(F)}\subseteq f^{-1}(F)$. Since the converse always holds, we have $f^{-1}(F)=\overline {f^{-1}(F)}$, so this set is closed and $f$ is continuous.
 A: You're making it way more complicated that it need to be and you also mix global continuity with local continuity...
We have the condition 
$$\forall V \in \mathcal{T}_Y: f(x) \in V \implies x \notin \overline{f^{-1}[V]^\complement}\tag{*}$$
If $f$ is continuous at $x$ then (*) holds: suppose $V$ is open in $Y$ containing $f(x)$, then we have $U$ open in $X$, containing $x$, such that $f[U] \subseteq V$. So $U \cap f^{-1}[V]^\complement = \emptyset$ and this neighbourhood $U$ witnesses that $x \notin \overline{f^{-1}[V]^\complement}$
If (*) holds, $f$ is continuous at $x$: suppose $V$ is an open neighbourhood of $f(x)$. We know by the condition that $x \notin \overline{f^{-1}[V]^\complement}$, so for some open neighbourhood $U$ of $x$ we have that $U \cap f^{-1}[V]^\complement = \emptyset$, or equivalently $f[U] \subseteq V$, as required. having this $U$ for our $V$ makes $f$ continuous at $x$.
So the condition (*) is seen to be a simple reformulation of the local continuity condition at $x$:
$$\forall V \in \mathcal{T}_Y: f(x) \in V \implies (\exists U \in \mathcal{T}_X: x \in U \land f[U] \subseteq V)\tag{c}$$ 
once you realise that $x$ is in the closure of a set iff every open neighbourhood of $x$ intersects that set and that $U \cap f^{-1}[V]^\complement = \emptyset$ and $f[U]\subseteq V$ are equivalent statements.
