Consider an orthonormal basis $\{x_j\}$ of $H$ with $x_1=x$, and consider $\{E_{kj}\}$ the corresponding matrix units (we don't really need matrix units, just the projection onto the span of $x$, but it might help understand). Assume that you can write $F_x=\alpha f + (1-\alpha)g$, for some $\alpha\in[0,1]$ and states $f,g$. Then, since $0\leq E_{11}\leq I$,
$$
1=\langle E_{11}x,x\rangle = F_x(E_{11})=\alpha f(E_{11}) + (1-\alpha) g(E_{11})\leq
\alpha+1-\alpha = 1.
$$
So $\alpha f(E_{11}) + (1-\alpha) g(E_{11})=1$. But as $f(E_{11})\leq1$, $g(E_{11})\leq1$, we conclude that $f(E_{11})=g(E_{11})=1$. In particular, $f(I-E_{11})=0$. Then, for any $T\in B(H)$,
$$
0\leq|f(T(I-E_{11}))|\leq f(T^*T)^{1/2} f((I-E_{11})^2)^{1/2}=f(T^*T)^{1/2}f(I-E_{11})^{1/2}=0.
$$
Thus $f(T)=f(T\,E_{11})$ for all $T$. Taking adjoints, $f(T)=f(E_{11}T)$. But then $f(T)=f(TE_{11})=f(E_{11}TE_{11})$. As $E_{11}TE_{11}=\langle Tx,x\rangle\,E_{11}=F_x(T)E_{11}$,
$$
f(T)=f(E_{11}TE_{11})=F_x(T)\,f(E_{11})=F_x(T).
$$
Similarly with $g$, so $F_x$ is an extreme point.
Edit: thanks to Matthew for noting that my argument in the last paragraph from the previous version was wrong. Here is the new argument.
Let $\pi_0:B(H)\to B(H)/K(H)$ be the quotient map onto the Calkin algebra $C(H)$. Let $\pi_1:C(H)\to B(K)$ be an irreducible representation of $C(H)$. Then
$$
\pi=\pi_1\circ\pi_0:B(H)\to B(K)
$$
is an irreducible representation. Using the correspondence between irreps and pure states, there exists a pure state $\varphi$ on $B(H)$ such that its GNS representation is unitarily equivalent to $\pi$. But this tells us that $\varphi(T)=0$ for all $T\in K(H)$, and so $\varphi$ cannot be a point state.
If $A=A^{\star}$ is a densely-defined selfadjoint linear operator on a complex Hilbert space $H$, and if there is a complete orthonormal basis of $H$ consisting of eigenvectors of $A$, then it is true that the point spectrum $\sigma_{p}(A)$ of $A$ is dense in $\sigma(A)$. The converse is not true.
To show this, let $H$ be a Complex Hilbert space and suppose $A:\mathcal{D}(A)\subseteq H\rightarrow H$ is a densely-defined selfadjoint linear operator. Suppose $H$ has an orthonormal basis $\{ e_{\alpha} \}_{\alpha \in \Lambda}$ of eigenvectors of $A$ with corresponding eigenvalues $\{\lambda_{\alpha}\}_{\alpha\in\Lambda}$, which must be real. If $x\in\mathcal{D}(A)$, then
$$
Ax = \sum_{\alpha\in\Lambda}(Ax,e_{\alpha})e_{\alpha}=\sum_{\alpha\in\Lambda}(x,Ae_{\alpha})e_{\alpha}
= \sum_{\alpha\in\Lambda}\lambda_{\alpha}(x,e_{\alpha})e_{\alpha}.
$$
If $\mu \in \sigma_{p}(A)$, then $Ax=\mu x$ for some $0\ne x \in \mathcal{D}(A)$, which gives
$$
0=(\mu x- Ax)=\sum_{\alpha\in\Lambda}(\mu-\lambda_{\alpha})(x,e_{\alpha})e_{\alpha}.
$$
Because $x \ne 0$, then $(x,e_{\alpha})\ne 0$ for some $\alpha\in\Lambda$, which forces $\mu=\lambda_{\alpha}$. So
$\sigma_{p}(A)\subseteq\{ \lambda_{\alpha} : \alpha \in \Lambda\}$, while the opposite inclusion is obvious. Hence, $\sigma_{p}(A)=\{\lambda_{\alpha} : \alpha\in\Lambda\}$.
The spectrum $\sigma(A)$ is closed; so $\sigma_{p}(A)^{c}\subseteq \sigma(A)$ (here 'c' denotes topological closure.) To see that $\sigma_{p}(A)^{c} =\sigma(A)$, we assume $\lambda \notin\sigma_{p}(A)^{c}$ and show that $\lambda\notin\sigma(A)$. To show this, note that if $\lambda\notin\sigma_{p}(A)^{c}$, then there exists $\delta > 0$ such that
$$
|\lambda_{\alpha}-\lambda| \ge \delta > 0,\;\;\; \alpha\in\Lambda.
$$
Therefore, if $x \in \mathcal{D}(A)$,
$$
\|(A-\lambda I)x\|^{2}=\sum_{\alpha}|\lambda_{\alpha}-\lambda|^{2}|(x,e_{\alpha})|^{2} \ge \delta^{2}\|x\|^{2}.
$$
Because $A=A^{\star}$, the above is enough to prove that $\lambda\notin\sigma(A)$, which proves that $\sigma_{p}(A)^{c}=\sigma(A)$.
To see that the converse is false, let $H=L^{2}[0,1]\times L^{2}[0,1]$ and let $\{ e_{n}\}_{n=1}^{\infty}$ be a complete orthonormal subset of $L^{2}[0,1]$. Define
$$
A(f,g) = (tf(t),\sum_{n=1}^{\infty}q_{n}(g,e_{n})e_{n}),
$$
where $\{ q_{n}\}$ is an enumeration of the rational numbers in $[0,1]$. Then
$$
A(0,e_{n})=q_{n}(0,e_{n}),
$$
which implies that $\{ q_{n}\}\subseteq \sigma_{p}(A)$. And $A(f,g)=\lambda (f,g)$ implies $f=0$ and $g=e_{n}$ for some $n$. So, even though $\sigma_{p}(A)^{c}=[0,1]=\sigma(A)$, $H$ cannot have a basis of eigenvectors for $A$.
Best Answer
As user953376 pointed out in their comment one-dimensional subspaces are determined by a unit vector. To fill in some detail: as an application of the Riesz representation theorem for every bounded linear operator $T$ on $\mathscr H$ the image of which is a one-dimensional, there exist $\psi,\phi\in\mathscr H$ such that $T=|\phi\rangle\langle\psi|$ (that is, $T(z)=\langle\psi,z\rangle\phi$ for all $z\in\mathscr H$). Now if $T$ is a self-adjoint projection then $T=|\psi\rangle\langle\psi|$ and $\langle\psi,\psi\rangle=1$. Thus every pure state in the above sense is determined by a unit vector, giving rise to the mapping \begin{align*} \iota:\{\psi\in\mathscr H:\langle\psi,\psi\rangle=1\}&\to\{T\in\mathcal B(\mathscr H):T\text{ orthogonal projection},\operatorname{dim}(\operatorname{ran}(T))=1\}\\ \psi&\mapsto|\psi\rangle\langle\psi|\,. \end{align*} While $\Psi$ is surjective, it fails to be injective (again for the reason user953376 described in their comment): $\psi$ and $e^{i\alpha}\psi$ are mapped to the same operator under $\iota$. This is solved by replacing the domain of $\iota$ by equivalence classes $[\psi]$ under the equivalence relation $\psi\sim\psi':\Leftrightarrow \psi=\psi' e^{i\alpha}$ for some $\alpha\in\mathbb R$---then $\iota$ is the isomorphism you were looking for.
As for your second question: there are two main reasons to consider states to be positive semi-definite trace-class operators of unit trace---or, equivalently, positive normal functionals $f$ on a $W^*$-algebra such that $f({\bf1})=1$---as opposed to mere elements in a Hilbert space:
To expand on this second point: every state $M$ (positive semi-definite trace class operator of unit trace) on an arbitrary Hilbert space allows for a spectral decomposition (because trace class) $M=\sum_{j=1}^\infty p_j|\psi_j\rangle\langle\psi_j|$ with "probabilities" $p_j\in[0,1]$ (positivity), $\sum_{j=1}^\infty p_i=1$ (trace condition) and an orthonormal system $\{\psi_j:j\in\mathbb N\}$ in $\mathscr H$. Therefore the pure states are fundamental to the concept of states as those are the "building blocks", but in order to get the full picture one has to allow for mixtures of such states, which are mathematically described by such convex combinations.