I'm trying to understand an implication in a theorem. Suppose you're given topological spaces $(X,\mathcal{T})$ and $(Y,\mathcal{U})$ and $f\colon X\to Y$. Then if for every convergent filter base $\mathcal{F}\to x$ in $X$, $f[[\mathcal{F}]]\to f(x)$ in $Y$, then $f$ is continuous.
$f[[\mathcal{F}]]$ is the set of all direct images of $f$ on the sets $A$ in $\mathcal{F}$. The proof says take any $U\in\mathcal{U}$ and $x\in f^{-1}(U)$. The filter $\mathcal{F}$ of all neighborhoods of $x$ converges to $x$, so $f[[\mathcal{F}]]\to f(x)$. For some neighborhood $V$ of $x$, $f[V]\subset U$, so $V\subset f^{-1}(U)$, and $f^{-1}(U)\in\mathcal{T}$.
I don't understand the last part. Why would $V\subset f^{-1}(U)$ imply that $f^{-1}(U)$ is an open set in $X$?