Continuity in Vietoris topology

Let $a\in \mathbb{R}$ and let a function $f_a\ :\ [0,1]\to \mathcal{H}(\mathbb{R})$ be given by $$f_a(x)=[a,a+x].$$ I have to show that $f_a$ is continuous.

Definition: Let $(X,\tau)$ be a topological space and $\mathcal{H}(X)$ be the set of all non-empty compact subsets of $X$. For $V,U_0,U_1,U_2,\dots,U_n\in \tau$ let $$\mathscr{B}_X^\mathcal{H}(V,U_0,U_1,U_2,\dots,U_n)=\{K\in \mathcal{H}(X)\ :\ K\subseteq V\& \ K\cap U_i\neq \emptyset,\ i=0,1,\dots,n \}.$$ The topology generated by the above basis is called the Vietoris Topology.

I am unable to prove that $f_a$ is continuous.

$$\left\{\mathscr{B}_X^{\mathcal{H}}(X,U):U\in\tau\right\}\cup\left\{\mathscr{B}_X^{\mathcal{H}}(V):V\in\tau\right\}$$
is a subbase for the Vietoris topology. Thus, it suffices to show that $f_a^{-1}\left[\mathscr{B}_X^{\mathcal{H}}(X,U)\right]$ and $f_a^{-1}\left[\mathscr{B}_X^{\mathcal{H}}(U)\right]$ are open in $[0,1]$ for each $U\in\tau$.
• $x\in f_a^{-1}\left[\mathscr{B}_X^{\mathcal{H}}(X,U)\right]$ if and only if $[x,a+x]\cap U\ne\varnothing$, so you want to show that $$\{x\in[0,1]:[x,a+x]\cap U\ne\varnothing$$ is open for each $U\in\tau. •$x\in f_a^{-1}\left[\mathscr{B}_X^{\mathcal{H}}(U)\right]$if and only if$[x,a+x]\subseteq U$, so you want to show that $$\{x\in[0,1]:[x,a+x]\subseteq U\}$$ is open for each$U\in\tau\$.