I don't follow all the steps in a theorem regarding continuous functionals Hi: I'm reading another document on Hilbert spaces  and progressing some but I've come to a theorem in this document, many of  the steps of which I don't follow. 
I will state the theorem below and then explain which steps are not clear to me.
Theorem 5.1. Let $H$ be a hilbert space and $\phi: H \rightarrow \mathbb{C}$, be a continuous functional. Then, 
$\dim N(\phi)^\perp = 1$.
where $N(\phi) = \{h \in H| \phi(h) = 0\}$.
Proof: It is easy to see that $N(\phi)$ is a closed subspace of $H$. 
Since, $\phi \neq 0, N(\phi) \neq H$. 
Then, $N(\phi) \neq \{0\}$. ( recall that $H = N(\phi) \oplus N(\phi)^\perp$). 
Take $x_1 \neq 0, x_2 \neq  0$ in $N(\phi)^{\perp}$.
Then, there exists a complex number $a \neq 0$ such that $\phi(x_1) + a \phi(x_{2}) = 0$. 
Then, $\phi(x_1 + a x_{2} ) = 0$ which  implies that $x_1 + a x_{2} \in N(\phi) \oplus N(\phi).^{\perp} = \{0\}$. 
Thus, $\dim N(\phi)^\perp = 1$. 
END OF THEOREM.
There are many steps in above that I don't follow.
1) Then, $N(\phi) \neq \{0\}$. I don't see where that comes from and does that indicate the null set or zero ?
2) $\phi(x_1) + a \phi(x_{2}) = 0$. Is this because the linear combo of  2 objects in a hilbert space is also in that space ?
3) $\phi(x_1 + a x_{2} ) = 0$. This is clearly crucial to the proof. I don't get it.
4) implies that $x_1 + a x_{2} \in N(\phi) \oplus N(\phi).^{\perp} = \{0\}$. Why does that imply it and again, is that the null set or zero ?
5) $\dim N(\phi)^\perp = 1$.  Don't follow this either ?
Thanks a lot.
 A: (1) It seems from context that there is a typo and they meant to write $N(\phi)^\perp \ne \{0\}$.  That is, $N(\phi)^\perp$ is not equal to the set containing only the zero vector.  Since every linear subspace contains the zero vector, this means $N(\phi)^\perp$ must contain some elements other than the zero vector, which justifies the following line.
(2) Note that $\phi(x_1)$ and $\phi(x_2)$ are just (nonzero) complex numbers.  Using simple algebra, you can very easily find what $a$ should be in order to make $\phi(x_1) + a \phi(x_2) = 0$.
(3) Recall that $\phi$ is a linear functional.
(4) I think they meant to write $x_1 + a x_2 \in N(\phi) \cap N(\phi)^\perp$.  We have just shown that $\phi(x_1 + a x_2) = 0$, which is the very definition of $x_1 + a x_2 \in N(\phi)$.  And we started out assuming that $x_1, x_2 \in N(\phi)^\perp$, and $N(\phi)^\perp$ is a linear subspace... Finally, recall that a set and its orthogonal complement always have only the zero vector in common.  (Any vector common to both must be orthogonal to itself.)
(5) We have just shown that $x_1 + a x_2 = 0$ for arbitrary $x_1, x_2$, which is to say that any two vectors in $N(\phi)^\perp$ are linearly dependent.  If the dimension of $N(\phi)^\perp$ was at least 2, we would be able to find two linearly independent vectors, and we've just shown that can't be done.
A: The other answer covers it, but here is another approach, too long for a comment:
$H=N(\phi)\oplus N(\phi)^\perp$ (because $H$ is a Hilbert space) and $H/N(\phi)\cong \mathbb C$ (By first isomorphism theorem), so the dimension of the quotient is $1$.
There is an $h_1\in H$ such that $h_1+N(\phi)$ generates $H/N(\phi)$. Now, $h_1=h+k$ for some $ h\in N(\phi),\ k\in N(\phi)^{\perp},$ so $h_1+N(\phi)=k+N(\phi).$ 
Therefore, if $x\in N(\phi)^{\perp},$  then $[x]=a(k+N(\phi))$ for some $a\in \mathbb C$, from which it follows that $x-ak\in N(\phi)$, but since each of these terms is contained in $N(\phi)^{\perp},$ we must have $x-ak=0$, and so $x=ak$. The result follows. 
