Continuity characterized via filterbases suppose that $f$ is not continuous at $x$. Consider a filterbase $\mathcal{B}=\{B_i \mid i \in I\}$. Since$f$ is not continuous at $x$, there exists an open neighbourhood $V(f(x)) \in \mathcal{T}_Y$ such that $f(B) \cap [Y-V(f(x))] \neq \emptyset$ for every $B \in \mathcal{B}$. Let $y \in f(B_i) \cap [Y-V(f(x))]$. Since $y \in f(B_i)$ there exists, $x \in B_i$ such that $y= f(x)$. We show that $\mathcal{B}$ in $X$ converges to $x$ but $f(\mathcal{B})$ in $Y$ does not converge to $f(x)$. How do I do this last part?
 A: I’m going to assume that you’re trying to prove that if $f:X\to Y$ is not continuous at some $x\in X$, then there is a filter base $\mathscr{B}$ in $X$ such that $\mathscr{B}\to x$, but $\{f[B]:B\in\mathscr{B}\}\not\to f(x)$.
Since $f$ is not continuous at $x$, there is an open nbhd $U$ of $f(x)$ such that whenever $V$ is an open nbhd of $x$ in $X$, $f[V]\nsubseteq U$. Let $\mathscr{B}$ be the family of all open nbhds of $x$. Clearly $\mathscr{B}\to x$, and for all $B\in\mathscr{B}$ we have $f[B]\nsubseteq U$, so $\{f[B]:B\in\mathscr{B}\}\not\to f(x)$.
It’s possible, though, that you’re trying to prove the opposite implication: if there is a filter base $\mathscr{B}$ in $X$ such that $\mathscr{B}\to x$ and $\{f[B]:B\in\mathscr{B}\}\not\to f(x)$, then $f$ is not continuous at $x$. 
If that’s what you want, assume that there is a filter base $\mathscr{B}$ in $X$ such that $\mathscr{B}\to x$ and $\{f[B]:B\in\mathscr{B}\}\not\to f(x)$. Then there is an open nbhd $U$ of $f(x)$ such that $f[B]\nsubseteq U$ for all $B\in\mathscr{B}$, and hence $f[B]\setminus U\ne\varnothing$ for all $B\in\mathscr{B}$. Now let $V$ be any open nbhd of $x$. $\mathscr{B}\to x$, so there is a $B\in\mathscr{B}$ such that $B\subseteq V$. Clearly $f[B]\subseteq f[U]$, so $f[V]\setminus U\supseteq f[B]\setminus U\ne\varnothing$, and $f[V]\nsubseteq U$. Thus, $x$ has no open nbhd $V$ such that $f[V]\subseteq U$, and therefore $f$ is not continuous at $x$.
