I want to show that given an alphabet $A = \{ L, R \}$, the language $$ \mathcal{L} = \{ x_{1} \ldots x_{n} \in A^{*} : \# \{ j \leq n : x_{j} = L \} = \# \{ j \leq n : x_{j} = R \} \}$$ cannot be expressed as the accepted language of an FDA. To see this, assume to the contrary that there is an FDA with states $1, \ldots, N$ that accepts exactly $\mathcal{L}$. Consider the words $L^{1}, \ldots, L^{N + 1}$. Then by Dirichlet, there must exist $i, j \leq N + 1, i < j$ such that $L^{i}$ and $L^{j}$ go to the same state. But if $L^{i}$ and $L^{j}$ are in the same state, and a word $w \in A^{*}$, then $L^{i}w$ and $L^{j}w$ will be in the same state. Let $w = R^{i}$. Then if $L^{i}R^{i} \in \mathcal{L}$ is in an acceptable state, then so must be $L^{j}R^{i} \not \in \mathcal{L}$; however, if $L^{i}R^{i}$ is not in an acceptable state, then the FDA has disallowed a word in the language. So either the FDA has "let through" $L^{j}R^{i}$, or "kept out" $L^{i}R^{i}$.
Is the above proof correct?