Number of real embeddings $K\to\overline{\mathbb Q}$ 
How many real embeddings, $K\to\overline{\mathbb Q}$ with $K=\mathbb Q\left(\sqrt{1+\sqrt{2}}\right)$ are there ? 

We set $f(x)=x^4-2x^2-1$ and if $\alpha=\sqrt{1+\sqrt{2}}$ then $f(\alpha)=0$.
Hence $\mathbb Q\left(\sqrt{1+\sqrt{2}}\right)=\mathbb Q[x]\big/(f(x))$
and the mappings $x\mapsto\pm\sqrt{1+\sqrt{2}}$ are real embeddings
and $x\to\pm\left(i \sqrt{\sqrt{2}-1}\right)$ are the complex embeddings
Is this approach true ?
Edit: How can I show that (without wolframalpha) $x^4-2x^2-1$ has $2$ pure imaginary roots ?

We note that an embedding is a non-trivial homomorphism (therefore injective as the domain is a field)
$$\varphi:\Bbb Q[x]/((x^2-1)^2-2)\to\Bbb C.$$
Where we call it "real" if the image is contained in $\Bbb R$. However, if $\beta$ denotes a choice of root of $f(x)=(x^2-1)^2-2$, then as

$$\Bbb Q(\beta)=\{a+b\beta+c\beta^2+d\beta^3 : a,b,c,d\in\Bbb Q\}$$

we see that this is a set of real numbers iff $\beta\in\Bbb R$, so that the number of ways to map our field into $\Bbb R$, i.e. the number of real embeddings, is exactly the number of real roots of $f(x)$.
Factoring we see

$$f(x)=(x^2-1-\sqrt{2})(x^2-1+\sqrt{2}).$$

Clearly the first factor has two real roots, and they are exactly as you found them, $\pm\sqrt{1+\sqrt{2}}$, and the second has none--its roots are the other you found, $\pm i\sqrt{\sqrt{2}-1}$--so the total number of real roots is $2$, and therefore the number of real embeddings is $2$.
