I don't know if it's true but I feel it should be. I want to prove:
Let $F:[0,1]\times X \rightarrow X$ be continuous. Call $F_a=F(a,\cdot).$ Let $K,V\subseteq X$ be compact and open respectively. Then, if $F_a(K)\subseteq V$, there exists an open interval $(a_1,a_2)$ around $a$ such that $\forall x\in (a_1,a_2)$ we have $F_x(K)\subseteq V$.
Thank you for your time.
On
Since $F$ is continuous $L:=F_a(K) $ is compact. If you know that you can separate two closed sets (in your question: $L$ and $B:=X\backslash V$) by an open set you are safe.
This is true, for example, if $X $ is a normal space (see this page about separation axioms: https://en.wikipedia.org/wiki/Separation_axiom). Weaker conditions may suffice.
If you can separate the sets, then you can find an open neighbourhood $U$ of $L$ in $B $, and you can find your interval by continuity of $F$ in the preimage of $U$.
By assumption, we have the inclusion,$$ \{ a \} \times K \subseteq F^{-1}(V),$$ Furthermore, $F$ is continuous, so $F^{-1}(V)$ must be an open subset of $[0,1] \times K$.
Therefore, for every $x \in K$, there exists an $\epsilon_x > 0$ and an open neighbourhood $U_x \subset K$ of $x$ such that $$ (a - \epsilon_x , a + \epsilon_x ) \times U_x \ \subseteq \ F^{-1}(V).$$ [This is obvious if you think about how the product topology is defined as the topology generated by products of open rectangles.]
Now then, observe that the family of sets $$ \{ U_x : x \in K \}$$ forms an open cover of $K$.
But since $K$ is compact, we can find a finite subcollection, $$ U_{x_1}, \ \ U_{x_2}, \ \ \dots \ \ U_{x_n},$$ that still covers $K$.
Thus, if we define $$ \epsilon = \min \{ \epsilon_{x_1}, \epsilon_{x_2}, \dots, \epsilon_{x_n}\}, $$ then $\epsilon$ is positive (because $\epsilon$ is the minimum of finitely many positive numbers), and we have the inclusion $$ (a - \epsilon, a + \epsilon) \times K \ \subseteq \ F^{-1}(V),$$ which is to say that $F_t(K) \subseteq V$ for all $t \in (a - \epsilon, a + \epsilon)$.