$T$ and $A$ are compact linear operators, $(T+\frac{1}{n} A)$ is also a compact linear operator for each $ n \in \mathbb{N}. $ Since $\mathbb{X}$ is reflexive, $(T+\frac{1}{n} A)$ attains norm for each $n \in \mathbb{N}$.\
Therefore, for each $n\in \mathbb{N},$ there exists $x_n\in S_\mathbb{X}$ such that $\|(T+ \frac{1}{n}A)x_n\|=\|T+ \frac{1}{n}A\|$.\
Since $\mathbb{X}$ is reflexive, $B_\mathbb{X}$ is weakly compact. Therefore $\{x_n\}$ has a weakly convergent subsequence. Without loss of generality we may assume that $\{x_n\}$ weakly converges to $x$ (say). Since $T$ and $A$ are compact linear operators, $Tx_n\rightarrow Tx$ and $Ax_n\rightarrow Ax$. Since $T\bot_B A$, $\|T+ \frac{1}{n}A\|\geq \|T\|$ for all $n\in \mathbb{N}$.\
Therefore, $\|(T+ \frac{1}{n}A)x_n\|=\|T+ \frac{1}{n}A\|\geq \|T\|\geq \|Tx_n\|$ for all $n\in \mathbb{N}$. Letting $n\rightarrow \infty$, we see that, $\|Tx\|\geq \|T\|\geq \|Tx\|$. This proves that $\|Tx\|=\|T\|,$ i.e., $x\in M_T$.\
Now we show that $Ax\in (Tx)^+$.\
For any $\lambda \geq \frac{1}{n},$ we claim that $\|Tx_n+ \lambda Ax_n\|\geq \|Tx_n\|$.\
Otherwise, $Tx_n+ \frac{1}{n} Ax_n= (1- \frac{1}{n\lambda}) Tx_n+ \frac{1}{n\lambda}(Tx_n+ \lambda Ax_n)$ gives that,\
$ \|T+\frac{1}{n}A\|=\|Tx_n+ \frac{1}{n} Ax_n\|\leq (1- \frac{1}{n\lambda}) \|Tx_n\|+ \frac{1}{n\lambda}\|(Tx_n+ \lambda Ax_n)\|<(1- \frac{1}{n\lambda}) \|Tx_n\|+ \frac{1}{n\lambda}\|Tx_n\|= \|Tx_n\| \leq \|T\|$, a contradiction. This completes the proof of our claim.\
Now for any $\lambda> 0$, there exists $n_0 \in \mathbb{N}$ such that $\lambda>\frac{1}{n_0}$. So for all $n\geq n_0$, 