If $C$ is symmetric and $\operatorname{rank}C=1$, then there is a $u\in\mathbb R^n$ and a $\alpha\in\mathbb R$ such that $C=\alpha uu^T$ Let $C\in\mathbb R^{n\times n}$. I want to show, that


If $0\ne C=uv^T$ for some $u,v\in\mathbb R^n$, then $$\operatorname{rank}C=1$$
If $C$ is symmetric and $\operatorname{rank}C=1$, then there is a $u\in\mathbb R^n$ and a $\alpha\in\mathbb R$ such that $$C=\alpha uu^T$$


(1.) is easy, since $$C=(v_1u,\ldots,v_nu)\;,$$ $\operatorname{rank}C$ is at most $1$. Since $C\ne 0$, the statement follows.

However, I don't know what I need to do for (2.). Clearly, $$\operatorname{rank}(\alpha uu^T)=1\;,$$ by $(1.)$. So, maybe we should assume that $$C=\alpha uu^T+A\;,$$ for some $A\in\mathbb R^n$, and show, that this leads to a contradiction to $\operatorname{rank}C=1$. Can we do this? The inequality $$\operatorname{rank}(X+Y)\le\operatorname{rank}X+\operatorname{rank}Y$$ doesn't seem to help.
Since $C$ has rank $1$, the column space of $C$ has dimension $1$, which means every column of $C$ is a multiple of some vector $u$. Hence
$$C=[v_1 \vec{u}, \cdots , v_n\vec{u}]=\vec{u}\vec{v}^T$$
for some vector $\vec{v}=(v_1, \dots, v_n)^T$.

Now if $C$ is symmetric, then $C=C^T=\vec{v}\vec{u}^T$. This means every column of $C$ is also a multiple of $\vec{v}$. Hence $\vec{u}$ is a multiple of $\vec{v}$, i.e.,

$$\vec{u}=k\vec{v}$$

You can see the result follows straightforwardly.
considering $C= uu^T + A$ doen't solve the problem because there might be a matrix $\lambda$ so that $\alpha uu^T + A = \beta\lambda \lambda^T$ so $C$ again can be written as  $\beta\lambda \lambda^T$.
It seems that you have already know what matrix $u$ is!