Is there an interval in the intersection of compact subsets under such a condition?

Let $\mathcal{F}=\{F_a\}_{a\in A}$ be a family of (arbitrary) subsets of $\mathbb R$ such that, for each countable set $D\subseteq A,$ the intersection $\bigcap_{a\in D}F_a$ contains a closed interval of length $1,$ and assume for a contradiction that $\bigcap_{a\in A}F_a$ contains no closed interval of length $1.$

Let $\{I_n:n\in\mathbb N\}$ be the set of all intervals of length $\ge1$ with rational endpoints. For each $n\in\mathbb N$ choose an index $b_n\in A$ such that $I_n\not\subseteq F_{b_n}.$ Let $B=\{b_n:n\in\mathbb N\}$ and let $F_B=\bigcap_{a\in B}F_a.$

The set $F_B$ contains no rational interval of length $\ge1,$ and no interval at all of length $\gt1.$ Since $B$ is countable, the set $F_B$ contains one or more closed intervals of length exactly $1,$ but these intervals must be pairwise disjoint, so there are at most countably many of them.

Let $\{J_n:n\in\mathbb N\}$ be the set of all closed intervals of length $1$ contained in $F_B.$ For each $n\in\mathbb N,$ choose an index $c_n\in A$ such that $J_n\not\subseteq F_{c_n}.$ Let $C=\{c_n:n\in\mathbb N\}.$ Then $F_{B\cup C}=\bigcap_{a\in B\cup C}F_a$ contains no closed interval of length $1.$ Since $B\cup C$ is countable, we have arrived at a contradiction.