If $Ax \in \langle x \rangle=\{ax:a\in \mathbb{R}\}$ for every vector $x$, then $A$ is square and diagonal. I'm asked to prove or disprove the title statement. I'm looking for verification/critique of my proof.
This is false. This statement is claiming that if every vector $x$ is an eigenvector for $A$ corresponding to a real eigenvalue, then $A$ must be square and diagonal. Symbolically, $\forall x \exists a\in\mathbb{R}$ s.t. $(A-aI)x=0\Rightarrow A$ is square and diagonal.
It is true that $A$ is square, as $A-aI$ is only defined for square $A$. However, it is not required that $A$ be diagonal. Consider, for counterexample, $A_{2\times2}=J_{2\times2}$, the $2\times2$ matrix of all 1's, and let $x=[x_1,x_2]^T$. Solving the system of equations $(J-aI)x=0$, we get that $-ax_1+ax_2=0$. Thus, $A_{2\times2}=J_{2\times2}$ has every vector $x$ as an eigenvector corresponding to the eigenvalue 0.
 A: The statement is in fact true, in fact, more is true, $A=\lambda I$ for some $\lambda\in\newcommand{\RR}{\mathbb{R}}\RR$. 
Proof: If every nonzero vector is an eigenvector of $A$ for the same eigenvalue $\lambda$, then $(A-\lambda I)v=0$ for all $v$, so $A-\lambda I = 0$, and $A=\lambda I$. Thus it suffices to prove that if every nonzero vector is an eigenvector, they all have the same eigenvalue. We'll prove this by contradiction. Let $v$ and $w$ be eigenvectors for $A$ with different eigenvalues, $\lambda$ and $\mu$ respectively. Since $v+w\ne 0$ (otherwise $v=-w$ would have the same eigenvalue as $w$), $v+w$ is an eigenvector of $A$ with eigenvalue $\nu$ for some $\nu\in\RR$.
Then $\nu(v+w) = A(v+w) = \lambda v + \mu w$. Thus $(\nu-\lambda)v+(\nu-\mu)w = 0$. Since $v$ and $w$ are not linearly dependent (or one would be a scalar multiple of the other and they'd have the same eigenvalue), we have $\nu-\lambda=\nu-\mu=0$. But this implies that $\nu=\lambda=\mu$, contradicting the assumption that $\lambda\ne \mu$. 
Thus it's impossible for two eigenvectors of $A$ to have two different eigenvalues. Hence $A=\lambda I$ for some $\lambda\in\RR$.
Now why doesn't your counterexample work? Actually I'm confused as to why you think it works.
$$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}
\begin{pmatrix}2\\3\end{pmatrix}
=\begin{pmatrix}5 \\5\end{pmatrix}\ne \lambda\begin{pmatrix}2\\3\end{pmatrix}\text{ for any $\lambda$}.$$
A: We have already seen a proof of the stronger version of the statement by jgon. If you are just interested in the matrix being diagonal you could also do it the following way. You write $A=(a_1 \dots a_n)$ where $a_j\in \mathbb{R}^n$. Then for every $j\in \{ 1 , \dots, n \}$ there exists $c_j \in \mathbb{R}$ such that
$$ a_j = A e_j = c_j e_j $$
where $e_j$ is the $j$th standard basis vector. Hence, $A=diag(c_1, \dots, c_n)$, i.e. $A$ is diagonal.
