I am not really sure what you are asking in your question, but here is the complete transformation into Chomsky-Normal-Form. The provided solution seems to be wrong, if they are really trying to get rid of the $\varepsilon$.
Step 1. Eliminate $\varepsilon$-rules
Every occurrence of $T$ could be $\varepsilon$, so like you already stated in your question the result is the following. Note however that $\varepsilon$ itself is part of the language, so even though you are eliminating $\varepsilon$-rules, $\varepsilon$ itself should appear in the production rules.
$S \to abSab \mid baSba \mid TT \mid T \mid \varepsilon,\\
T \to aTa \mid aa \mid bTb \mid bb$
Step 2. Eliminate chain rules (rules of the form $A \to B$, $B \to \mathrm{expression}$)
$S \to abSab \mid baSba \mid TT \mid aTa \mid aa \mid bTb \mid bb \mid \varepsilon,\\
T \to aTa \mid aa \mid bTb \mid bb$
Step 3. Replace terminals (except if there is only one terminal on the right hand side)
$S \to T_aT_bST_aT_b \mid T_bT_aST_bT_a \mid TT \mid T_aTT_a \mid T_aT_a \mid T_bTT_b \mid T_bT_b \mid \varepsilon,\\
T \to T_aTT_a \mid T_aT_a \mid T_bTT_b \mid T_bT_b,\\
T_a \to a,\\
T_b \to b$
Step 4. Shorten rules
$S \to T_a \mid T_bH_2 \mid TT \mid T_aH_3 \mid T_aT_a \mid T_bH_4 \mid T_bT_b \mid \varepsilon,\\
T \to T_aH_3 \mid T_aT_a \mid T_bH_4 \mid T_bT_b,\\
T_a \to a,\\
T_b \to b,\\
H_1 \to T_bH_{11},\\
H_{11} \to SH_{12},\\
H_{12} \to T_aT_b,\\
H_2 \to T_aH_{21},\\
H_{21} \to SH_{22},\\
H_{22} \to T_bT_a,\\
H_3 \to TT_a,\\
H_4 \to TT_b\\
$
ε could be replaced by T
(with one a bit sloppy). What shall it mean? Does this appear in the original grammar? $\endgroup$ – greybeard Oct 7 '20 at 8:32