Let $x \in \Bbb R^n$ be a vector.
Claim: $xx^T$ is an extreme vector of positive semi-definite ($PSD_n$) matrices.
We have to show that if $xx^T = A + B$, where $A,B \in PSD_n$, then both $A,B$ are scalar multiple of $xx^T$.
$ns(A) \cap ns(B) = ns\{xx^T\} = \{x\}^{\perp}$, where ns means null-space.
So, either $ns(A) = \{x\}^{\perp} \text{or} \ \Bbb R^n$. In case of $ns(A) = \Bbb R^n$, we are done. I am having problem when $ns(A) = \{x\}^{\perp}$, how to conclude that $A$ is scalar multiple of $xx^T$.
Suppose that $ns(A) = \{x\}^\perp$. By the rank-nullity theorem, $A$ is a rank $1$ matrix. That is, we necessarily have $A = uv^T$ for some vectors $u,v$. Because $A$ is symmetric, $u$ is a multiple of $v$. Because $ns(A) = \{x\}^\perp$, $v$ must be a multiple of $x$.
Conclude that $A$ is a multiple of $xx^T$, as desired.