A recent question in this forum led me to recall the linear algebra result:
$$
\mbox{rank}(A + B) \leq \mbox{rank}(A) + \mbox{rank}(B), \ \ \left( A, B \in \mathbf{R}^{n \times n} \right)
$$
(I corrected my first posting after Ben Grossmann corrected it and cited an important inequality in linear algebra. I like to thank him first!)
Define $U = \mbox{Range}(A)$, $V = \mbox{Range}(B)$.
Then $U$ and $V$ are subspaces of $\mathbf{R}^n$.
Clearly, $\mbox{rank}(A) = \mbox{dim}(U)$, $\mbox{rank}(B) = \mbox{dim}(V)$.
Also, $\mbox{rank}(A + B) \leq \mbox{dim}(U + V)$.
(Thanks to Ben Grossmann for correcting my original statement!)
We know the theorem from linear algebra:
$$
\mbox{dim}(U + V) = \mbox{dim}(U) + \mbox{dim}(V) – \mbox{dim}(U \cap V)
$$
which implies that
$$
\mbox{dim}(U + V) \leq \mbox{dim}(U) + \mbox{dim}(V).
$$
Thus, $\mbox{rank}(A + B) \leq \mbox{dim}(U + V) \leq \mbox{rank}(A) + \mbox{rank}(B)$.
Best Answer
You have made one mistake: it is not necessarily the case that $\operatorname{rank}(A + B) = \dim(U + V)$. However, it is true that $\operatorname{Range}(A + B) \subseteq U + V$, which means that $\operatorname{rank}(A + B) \leq \dim(U + V)$.
Otherwise, your proof is correct and complete. After changing $\operatorname{rank}(A + B) = \dim(U + V)$ to $\operatorname{rank}(A + B) \leq \dim(U + V)$, the rest of your proof works as is.
A quick counterexample to $\operatorname{rank}(A + B) = \dim(U + V)$: consider $$ A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{-1&0\\0&-1}. $$