Spectral norm of projected matrix

This is not true. E.g. suppose that $n=m=2$ and $T$ is the orthogonal projection onto the linear span of $B=\operatorname{diag}(6,3)$. Let $A=5I$. Then $T(A)=B$ because $\langle A-B,B\rangle=\langle \operatorname{diag}(-1,2),\operatorname{diag}(6,3)\rangle=0$. However, $\|T(A)\|_2=\|B\|_2=6>5=\|A\|_2$.

Another counterexample: let \begin{aligned} &T(X)=X-\pmatrix{1\\ &0}X\pmatrix{1\\ &0},\\ &A=\pmatrix{-1&1\\ 1&1},\ B=T(A)=\pmatrix{0&1\\ 1&1}, \end{aligned} then $\|T(A)\|_2=\|B\|_2=\frac{1+\sqrt{5}}{2}\approx1.618>1.414\approx\sqrt{2}=\|A\|_2$.

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Remark. The inequality that motivated the OP's question, namely, $$ \|T(A)\|_F\le\sqrt{\operatorname{rank}(T(A))}\,\|A\|_2 $$ where $T(A)=A-P_1AP_2$ for some two orthogonal projectors $P_1$ and $P_2$, is correct. Without loss of generality, we may assume that $$ A=\pmatrix{X&Y\\ Z&W} \ \text{ and }\ T(A)=\pmatrix{0&Y\\ Z&W}. $$ Let $Y=U_y(S_y\oplus0)V_y^T$ and $Z=U_z(S_z\oplus0)V_z^T$ be two singular value decompositions, where $S_y$ and $S_z$ are two positive diagonal matrices. Define on $M_{n,m}$ a linear map $$ F:B\mapsto\pmatrix{U_y\\ &U_z}^TB\pmatrix{V_z\\ &V_y}. $$ Then we may write $$ F(A)= \left(\begin{array}{c&c|c&c}\ast&\ast&S_y&0\\ \ast&\ast&0&0\\ \hline S_z&0&E&F\\ 0&0&G&H\end{array}\right) \ \text{ and }\ F(T(A))= \left(\begin{array}{c&c|c&c}0&0&S_y&0\\ 0&0&0&0\\ \hline S_z&0&E&F\\ 0&0&G&H\end{array}\right). $$ Let $$ M_1=\pmatrix{S_y\\ 0\\ 0\\ G}\ \text{ and }\ M_2=\pmatrix{S_z&0&E&F}. $$ Since $M_1,M_2$ and $H$ are submatrices of $F(A)$, their spectral norms are bounded above by $\|F(A)\|_2$. It follows that \begin{aligned} \|T(A)\|_F^2 &=\|F(T(A))\|_F^2\\ &=\|M_1\|_F^2+\|M_2\|_F^2+\|H\|_F^2\\ &\le\operatorname{rank}(M_1)\|M_1\|_2^2+\operatorname{rank}(M_2)\|M_2\|_2^2+\operatorname{rank}(H)\|H\|_2^2\\ &\le\left(\operatorname{rank}(M_1)+\operatorname{rank}(M_2)+\operatorname{rank}(H)\right)\|F(A)\|_2^2\\ &=\operatorname{rank}(F(T(A)))\|F(A)\|_2^2\\ &=\operatorname{rank}(T(A))\|A\|_2^2. \end{aligned}