How to prove that EPR outcomes have equal probability no matter the basis?

Recently in class, we learned about the EPR state. I know that no matter what basis the first qubit is measured in, the two outcomes have an equal probability. However, how does one prove this? I somewhat understand what is happening and why, but if you know where a through proof of this might be that would be amazing.

Let $$\rho_{AB}$$ be a quantum state shared between two parties, Alice and Bob. Suppose Alice performs a POVM measurement $$\{M_i\}_i$$ on her half of the state. Then the probability that Alice obtains outcome $$i$$ is given by the Born rule as $$p(i) = \mathrm{Tr}[\rho_{AB}(M_i \otimes I)].$$ But whenever we have the trace of a multipartite operator we can always perform a partial trace first, i.e., $$\mathrm{Tr}[X] = \mathrm{Tr}[\mathrm{Tr}_B[X]]$$. So in this case $$p(i) = \mathrm{Tr}[\mathrm{Tr}_B[\rho_{AB}(M_i \otimes I)]] = \mathrm{Tr}[\rho_{A} M_i]$$ where $$\rho_A = \mathrm{Tr}_B[\rho_{AB}]$$. For the EPR states, if you calculate $$\rho_A$$ (you should verify this) you'll find it is equal to $$I/2$$ (the maximally mixed state). Thus for the EPR states $$p(i) = \mathrm{Tr}[I/2 M_i] = \frac12 \mathrm{Tr}[M_i].$$ Now if $$M_i$$ is a rank one qubit projector (which I'm assuming is what you meant by a measurement) then $$\mathrm{Tr}[M_i] = 1$$.
The local state (described by density matrix) of each qubit in EPR state is $$\begin{equation} \rho=\frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation}$$ It does not depend on basis, so both outcomes have 50% probabilities in every measurement basis.
The EPR state is pure 2-qubit state $$|\Psi\rangle_{EPR}= \frac{1}{\sqrt{2}}(|0\rangle|0\rangle+|1\rangle|1\rangle)$$ or in the standard basis $$|\Psi\rangle_{EPR}= \frac{1}{\sqrt{2}}\begin{pmatrix}{1\\0\\0\\1} \end{pmatrix}$$
The 2-qubit density matrix is $$\rho_{AB}=|\Psi\rangle_{EPR} \langle \Psi|_{EPR}= \frac{1}{\sqrt{2}}\begin{pmatrix}1\\0\\0\\1 \end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 0 & 0 & 1 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 1&0&0&1\\ 0&0&0&0\\ 0&0&0&0\\ 1&0&0&1 \end{pmatrix}$$ and the local state of say qubit $$A$$ is given by the partial trace over qubit $$B$$ $$\rho_A=tr_B[\rho_{AB}]=\frac{1}{2}\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}$$ The same expression is obtained for qubit $$B$$