The main point is that operations in subfields coincide with operations in the large field. That's the definition of subfield.
In the particular case of inverses, you use uniqueness:
Let $a\in \mathcal K = \bigcap_{i\in\mathcal I}\mathcal K_i$. Since $\mathcal K_i$ is field, there is $b_i \in \mathcal K_i$ such that $ab_i=1$. Since $\mathscr K$ is a field, there is $b \in \mathscr K$ such that $ab=1$. But then $b=b1=bab_i = 1b_i=b_i$. Thus, $b\in \mathcal K_i$ for all $i$ and so $b\in \mathcal K$.
The proof above works for arbitrary families; it does not depend on the family being countable.
I'll give you two different ways to do this as I don't know if you know the main theorem yet but even if you don't you can revisit this once you do learn it.
Method 1:
The simplest method is to use the Fundamental Theorem/ Galois Correspondence Theorem which says the intermediate fields are in bijection with subgroups of the Galois group, which is isomorphic to $C_2 \times C_2$ in this case.
This has $5$ subgroups so we have $5$ intermediate fields which are $\mathbb{Q}$, $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{5})$, $\mathbb{Q}(i\sqrt{5})$ and $\mathbb{Q}(i, \sqrt{5})$ which are easily spotted.
Method 2:
If you haven't got that far yet then we can use the tower law instead. Note $[\mathbb{Q}(i,\sqrt{5}):\mathbb{Q}]=4$ so any nontrivial subfield will have degree $2$.
It's a standard result that any quadratic field has the form $\mathbb{Q}(\sqrt{D})$ for some squarefree integer $D$. Using your basis $a+bi+c\sqrt{5} +di\sqrt{5}$, we can see $i=\sqrt{-1}$, $\sqrt{5}$ and $i\sqrt{5}=\sqrt{-5}$ all lie in $\mathbb{Q}(i,\sqrt{5})$ so all give quadratic subfields.
Now suppose $\mathbb{Q}(\sqrt{D})$ was also a subfield. Then$\sqrt{D} \in \mathbb{Q}(i,\sqrt{5})$. This means $\sqrt{D} = a+bi+c\sqrt{5} +di\sqrt{5}$ for some $a,b,c,d$.
Squaring both sides we get $D= (a^2 - b^2 +5c^2 - 5d^2) + (2ab+10cd)i + (2ac-2bd)\sqrt{5} +(2ad+2bc)i\sqrt{5}$.
We are then left with solving the simulataneous equations:
$\begin{eqnarray*}
D &=& a^2 - b^2 +5c^2 - 5d^2, \\
0 &=& 2ab+10cd, \\
0 &=& 2ac-2bd, \\
0 &=& 2ad+2bc,
\end{eqnarray*}$
which then gives you solutions only for $D=-1,5,-5$ (remembering that we only consider squarefree $D$), but this is quite tedious.
Best Answer
First note that $\mathbb Q$ is itself a subfield of $\mathbb R$, so the intersection of all subfields must be a subset of the rationals.
Second note that $\mathbb Q$ is a prime field, that is, it has no proper subfields. This is true because if $F\subseteq\mathbb Q$ is a field then $1\in F$, deduce that $\mathbb N\subseteq F$, from this deduce that $\mathbb Z\subseteq F$ and then the conclusion.
Third, conclude the equality.