my assumptions:
(i) $\lim_{t \to \infty}F_{t}(x)=F(x) \ \forall\ x\ \in\ C(F)$(set of continuity points of F) with $F_{t}(x)$ family of distribution functions and $F$ distribution function
(ii) $\lim_{t \to \infty}x_{t}=x_{0}$ ($x_{0}\in C(F)$)
this we can write as: $\forall\ \epsilon>0\ \exists T$ such that $|x_{t}-x_{0}|<\epsilon$ for $t>T$
i want to show that then also the following statement is valid:
$\lim_{t \to \infty}F_{t}(x_{t}) =F(x_{0})$
My proof
first i`m looking for an upper bound of $F_{t}(x_{t})$:
$F_{t}(x_{t})=P_{t}(X_{t}\leq x_{t})\leq P_{t}(X_{t}\leq \epsilon+x_{0}) $ for t sufficient large
with the help of this upper bound we know that there exists the limsup of $F_{t}(x_{t})$ such that $\limsup_{t \to \infty}F_{t}(x_{t})\leq F(x_{0}+\epsilon)$ an because of the right continuity of F and as we can choose $\epsilon$ very small we get $\limsup_{t \to \infty}F_{t}(x_{t})\leq F(x_{0})$
When i go the same way to get a lower bound i will get
$F(x_{0})\leq \liminf_{t \to \infty}F_{t}(x_{t})$ and as we know that both exist and valid the following unequality
$F(x_{0})\leq \liminf_{t \to \infty} F_{t}(x_{t})\leq \limsup_{t \to \infty}F_{t}(x_{t})\leq F(x_{0})$
they must be equal and so it is done.
second try (here i try an other way and i think on this way there is no problem with the missing left continuity)
$F$ continuity in $x_{0}$ then the following inequality is valid:
$F(x_{0})-\epsilon<F(x)<F(x_{0})+\epsilon \ \forall x \mbox{with}\|x-x_{0}|<\delta$ but then there exists also continuity points $x_{1},x_{2}$ mit $x_{1}<x_{0}<x_{2}$ such that
$F(x_{0})-\epsilon<F(x_{1})\leq F(x_{0})\leq F(x_{2})<F(x_{0})+\epsilon $ applying assumption (i) we get
$F(x_{0})-\epsilon<F_{t}(x_{1})\leq F_{t}(x_{0})\leq F_{t}(x_{2})<F(x_{0})+\epsilon $
And we know for t sufficiently large $x_{t} \in (x_{1},x_{2})$ so we get
$F(x_{0})-\epsilon<F_{t}(x_{1})\leq F_{t}(x_{t})\leq F_{t}(x_{2})<F(x_{0})+\epsilon $
Is it necessary to work with limsup and liminf? i think on this way it is not, because now we can choose $\epsilon$ small and get
$F_{t}(x_{0})\leq F_{t}(x_{t})\leq F_{t}(x_{0})$