It is true that $$\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)\subset \bigcap_{i\in I}\text{int}(A_i),$$ and the inclusion can be proper if the index set is infinite. Keeping our attention only on metric spaces, which are Hausdorff, if metric space $(X,d)$ has a non-isolated point $x$, the sets $A_i=\{y\in X:d(x,y)<1/i\}$ are individually open with $x\in \text{int}(A_i)=A_i$, but $\cap_{i=1}^\infty A_i=\{x\}$. As, by hypothesis, $x$ is not isolated, $\{x\}$ is not open, and therefore does not contain $x$ (or anything else) in its interior. Basically the same example can be found in any $T_1$ space with a non-isolated point.
This inclusion $$\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)\subset \bigcap_{i\in I}\text{int}(A_i)$$ is true in any topological space.
Since you are working in the real line, let's start there with that as an example.
If $A\subset \mathbb{R}$, then a point $x\in\mathbb{R}$ satisfies $x\in \text{int}(A)$ if and only if there exists $\delta>0$ such that $(x-\delta,x+\delta)\subset A$.
Therefore $x\in \text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)$ if and only if there exists $\delta>0$ such that $$(x-\delta,x+\delta)\subset \bigcap_{i\in I}A_i,$$ which happens if and only if there exists $\delta>0$ such that for all $i\in I$, $$(x-\delta,x+\delta)\subset A_i.$$
On the other hand, $x\in \bigcap_{i\in I}\text{int}(A_i)$ means that for each $i\in I$, $x\in \text{int}(A_i)$, which means there exists $\delta_i>0$ such that $(x-\delta_i,x+\delta_i)\subset A_i$. But in this case, the different $i$ can have different $\delta_i$.
So membership in $\bigcap_{i\in I}\text{int}(A_i)$ means that for each $i\in I$ there exists $\delta_i>0$ such that $(x-\delta_i,x+\delta_i)\subset A_i$, while membership in $\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)$ means there is one $\delta>0$ which will accomplish this same thing for all $i$ at the same time.
To get down to it, it is the distinction between "For all $i\in I$ there exists $\delta$ such that $\ldots$" and "There exists $\delta$ such that for all $i\in I$ $\ldots$".
If $I$ is finite, we can take $\delta=\min_{i\in I}\delta_i$, because a finite set of positive numbers has a minimum, and these two notions are equivalent. You already came up with examples to show the failure if $I$ is infinite with $A_i=[0,1+1/i[$. So for $x=1$, the best $\delta_i$ for $A_i$ is $\delta_i=1/i$, and there is not one $\delta>0$ which will work simultaneously for all $i$.
In general metric spaces, say $(X,d)$, this is true if we replace intervals $(x-\delta,x+\delta)$ with $$B_\delta(x)=\{y\in X:d(x,y)<\delta\}.$$ These are intervals in $\mathbb{R}$, disks in $\mathbb{R}^2$ (with Euclidean distance, although there are other interesting distances in $\mathbb{R}^2$), etc.
It's still true in general topological spaces. Membership in $\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)$ means there exists an open set $U$ such that for all $i\in I$, $$x\in U\subset A_i.$$ So a single open set $U$ works simultaneously for all $i$. But $x\in \bigcap_{i\in I}\text{int}(A_i)$ means that for all $i\in I$, there exists an open set $U_i$ such that $$x\in U_i\subset A_i.$$ This is exactly the same idea, that "For all $i\in I$, there exists $U\in \tau$ such that $\ldots$" is weaker than "There exists $U\in \tau$ such that for all $i\in I$, $\ldots$". Here, $\tau$ is our topology (set of open sets).
A few interesting things that we can get out of this:
We always have $$\overline{A}^c=\text{int}(A^c),\tag{$I$}$$ where the superscript $c$ denotes the set complement and $\overline{A}$ denotes the closure of a set. To see why $I$ holds, for $x\in X$, either every open set containing $x$ intersects $A$ (which is equivalent to $x\in \overline{A}$) or there exists an open set containing $x$ which doesn't intersect $A$ (which is equivalent to $x\in \text{int}(A^c)$).
Since $(A^c)^c=A$, $I$ is equivalent to $$\overline{A^c}=\text{int}(A)^c.$$
We also have DeMorgan's laws for arbitrary unions/intersections: $$\Bigl(\bigcap_{i\in I}A_i\Bigr)^c = \bigcup_{i\in I}A_i^c$$ $$\Bigl(\bigcup_{i\in I}A_i\Bigr)^c = \bigcap_{i\in I}A_i^c.$$
Also, if $A\subset B$, then $A^c\supset B^c$.
What can we do with these observations? Let $(B_i)_{i\in I}$ be a collection of subsets (imagine subsets of $\mathbb{R}$, if for concreteness). Let $A_i=B_i^c$ for all $I$. We know $$\Bigl[\text{int}\Bigl(\bigcap_{i\in I}A_i\Bigr)\Bigr]^c \supset \Bigl[\bigcap_{i\in I}\text{int}(A_i)\Bigr]^c.$$ Applying $\overline{A^c}=\text{int}(A)^c$ with $A=\bigcap_{i\in I}A_i$, we get $$\overline{\Bigl(\bigcap_{i\in I}A_i\Bigr)^c }\supset \Bigl[\bigcap_{i\in I}\text{int}(A_i)\Bigr]^c.$$
Using DeMorgan's laws, we get $$\overline{\bigcup_{i\in I}A_i^c}=\overline{\Bigl(\bigcap_{i\in I}A_i\Bigr)^c }\supset \Bigl[\bigcap_{i\in I}\text{int}(A_i)\Bigr]^c = \bigcup_{i\in I}\text{int}(A_i)^c.$$ Using $\overline{A^c}=\text{int}(A)^c$ again, and the fact that $A_i^c=B_i$, we get $$\overline{\bigcup_{i\in I}B_i}=\overline{\bigcup_{i\in I}A_i^c}\supset \bigcup_{i\in I}\text{int}(A_i)^c = \bigcup_{i\in I}\overline{A_i^c}=\bigcup_{i\in I}\overline{B_i}.$$
So now we also know something analogous about closures: $$\overline{\bigcup_{i\in I}B_i}\supset\bigcup_{i\in I}\overline{B_i}.$$
