We let $A$ and $B$ be symmetric matrices of rank one, where $A \neq kB$ for any $k$. We wish to show that the rank of $A + B$ is 2.
By subadditivity, we have that the rank of $A+B$ is at most 2. For contradiction, we say the rank is not 2. In the case of the rank of $A+B$ being $0$, we get that $A=-B$ which is a contradiction as $A \neq kB$.
I seem to be stuck on the case in which the rank of $A+B$ is $1$. Would utilizing the fact that rank one matrices can be represented as outer products be helpful?
As $A$ has rank one, we may assume $A=\begin{pmatrix}\lambda_1 v & \cdots &\lambda_n v\end{pmatrix}$ where $v=\begin{pmatrix} v_1 \\ \vdots \\ v_n\end{pmatrix}$ is a nonzero column vector. Then $$A=\begin{pmatrix} v_1 \\ \vdots \\ v_n\end{pmatrix} \begin{pmatrix} \lambda_1 & \cdots & \lambda_n\end{pmatrix}=(v_i\lambda_j)_{i,j=1}^n$$
As $A$ is symmetric, we have $v_i\lambda_j=v_j\lambda_i$. As $v$ is nonzero, there is $v_j\not=0$, hence $\lambda_i = \frac{\lambda_j}{v_j}v_i$, that is there is a constant $C_1$, such that $\begin{pmatrix} \lambda_1 & \cdots & \lambda_n \end{pmatrix}=C_1\begin{pmatrix} v_1 & \cdots & v_n \end{pmatrix}$. Hence we established $A$ has decomposition: $$A=C_1 vv^t$$
Similarly, $B$ also has decomposition: $$B=C_2 ww^t$$
Note that $C_1, C_2$ must be nonzero for $A, B$ to have rank one.
And $$A+B=C_1vv^t+C_2ww^t$$
Since $A\not=kB$, we have $v,w$ are linearly independent. Therefore there exists a dual set $\{v', w'\}$, such that $$v^tv'=w^tw'=1, v^tw'=w^tv'=0$$ Hence $(A+B)v'=C_1v$ and $(A+B)w'=C_2w$ are linearly independent, so $\text{rank}(A+B)=2$.
The proof has made no assumption about the base field. But this is not true for non-symmetric matrices: $$A=\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}, B=\begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix}$$