proving that $a + b \sqrt {2} + c \sqrt{3} + d \sqrt{6} $ is a subfield of $\mathbb{R}$

Regarding 1: Note that if we multiple your general element by its conjugate with respect to $\sqrt{3}$, $$ (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6}) \\ = a^2 + 2b^2 - 3c^2 -6d^2 + (2ab-6cd)\sqrt{2} \text{,} $$ which we can simplify further by multiplying with this new expression's conjugate with respect to $\sqrt{2}$, $a^2 + 2b^2 - 3c^2 -6d^2 - (2ab-6cd)\sqrt{2}$. We find $$ (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6})(a^2 + 2b^2 - 3c^2 -6d^2 - (2ab-6cd)\sqrt{2}) \\ = a^4-4 a^2 b^2-6 a^2 c^2-12 a^2 d^2+48 a b c d+4 b^4-12 b^2 c^2-24 b^2 d^2+9 c^4-36c^2 d^2+36 d^4 \text{.} $$ Now divide both sides by $(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})$ and $(a^4-4 a^2 b^2-6 a^2 c^2-12 a^2 d^2+48 a b c d+4 b^4-12 b^2 c^2-24 b^2 d^2+9 c^4-36c^2 d^2+36 d^4)$ to find an expression for the multiplicative inverse of $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$.

Notice that this is entirely analogous to what we do with complex numbers. For $a+b\mathrm{i}$, we multiply by its conjugate (with respect to $\mathrm{i}$), $$ (a + b\mathrm{i})(a - b\mathrm{i}) = a^2 + b^2 \text{.} $$ Then $\frac{a - b \mathrm{i}}{a^2 + b^2} = \frac{1}{a + b \mathrm{i}}$.

Regarding 2: What's the multiplicative identity in $\mathbb{R}$? Is it possible for the multiplicative identity of a subfield to be a different number? Suppose $s$ in a subfield satisfies $sx = x$ for every $x$ in that subfield, then that equation is also true for all those elements in the field. If we write $1 \in \mathbb{R}$ as the multiplicative identity in $\mathbb{R}$, then $sx = 1x$ is an equivalent equation in $\mathbb{R}$ and $(s-1)x = 0$. It is a familiar property of $\mathbb{R}$ that this forces either $s-1 = 0$ or $x = 0$. Since this holds for at least one non-zero $x$ (because subfields are fields, so they have at least two elements), it must be the case that $s-1 = 0$, so $s = 1$.

Regarding 3: Every element in any field (or subfield) has an additive inverse. For a field $F$ and an element $x \in F$, its additive inverse is $-x$. For your element $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$, its additive inverse is $-a -b\sqrt{2} -c\sqrt{3} -d\sqrt{6}$ for every choice of $a,b,c,d$. Even zero has an additive inverse. (It's zero.)

Here it is for $a+b\sqrt{2}$.
The multiplicative identity is $1$, or you could write it as $1+0\sqrt2$.
A subfield is the same as a field. You just have to check the sum, difference, product and ratio stays in the set.
To check the ratio stays in the set, I just do the multiplicative inverse. That and multiplication gives any ratio.
Suppose the inverse of $a+b\sqrt2$ is $c+d\sqrt2$. Then
$$(a+b\sqrt2)(c+d\sqrt2)=1\\ ac+2bd+(ad+bc)\sqrt2=1+0\sqrt2\\ ac+2bd=1,ad+bc=0$$ The last two equations can be written as a matrix equation $$\left[\begin{array}{cc}a&2b\\b&a\end{array}\right] \left[\begin{array}{c}c\\d\end{array}\right]= \left[\begin{array}{c}1\\0\end{array}\right]$$ The matrix's determinant is $a^2-2b^2$. If its determinant is zero for nonzero rationals $a$ and $b$, then $\sqrt2=a/b$ is rational. So the determinant is nonzero and you can invert the matrix to find $c$ and $d$.
You will have a $4×4$ matrix instead of a $2×2$.

I think that all will agree that the only problem is deciding that a number of form $a+b\sqrt2+c\sqrt3+d\sqrt6$ (with $a,b,c,d\in\Bbb Q$) has a reciprocal that’s again of this form. You don’t need to write out the reciprocal, just give an adequate argument that your number has one. Let’s do it this way: $$ \begin{align} \frac1{a+b\sqrt2+c\sqrt3+d\sqrt6}&=\frac1{(a+b\sqrt2\,)+(c+d\sqrt2\,)\sqrt3}\\ \\ &=\frac{(a+b\sqrt2\,)-(c+d\sqrt2\,)\sqrt3}{(a+b\sqrt2\,)^2-3(c+d\sqrt2\,)^2}\,, \end{align} $$ in which the numerator is of form $a'+b'\sqrt2+c'\sqrt3+d'\sqrt6$, and the denominator now has the form $A+B\sqrt2$, for suitable rational numbers $A$ and $B$. But you already know, from high-school, how to find $1/(A+B\sqrt2)$.

And there you are.