I was reading Sadri Hassani book in which the author proves that if an automorphism of a vector space leaves one summand of a direct sum invariant, then the other one is invariant as well. The proof is as follows:-
I cannot understand why you have $T(V_1\oplus V_2) = T(V_1)\oplus T(V_2)$. The statement also seems wrong to me as if you suppose a 2 dimensional space with the automorphism operator $T$ given by(in matrix notation):-
\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} The vector $V_1$ could be your x axis and $V_2$ be the y axis. Now, T leaves the x axis invariant but doesn't leave the y axis invariant. Where is my understanding wrong?
As a side note to my comment, the step you had issues with is correct, in fact: $T(V_1 \oplus V_2) = T(V) = V$ For every $v \in V$ we have that $v = T(w)$ and therefore $v = T(v_1+v_2) = T(v_1)+T(v_2)$ so $T(V_1) + T(V_2) = V$. To see that the sum is direct, if $v \in T(V_1) \cap T(V_2)$ then $v = T(v_1) = T(v_2)$ but then $T(v_1-v_2) = 0$ which would imply $v_1 = v_2$ and therefore $v_1=v_2=0 \Rightarrow v=0$.
The problem lies in the assumption that $V_1 \oplus V_2 = V_1 \oplus V_2' \Rightarrow V_2 = V_2'$, as a counterexample you can take the spaces you just mentioned, $V_1 = \{(x,0)|x \in \mathbb{R}\}$, $V_2 = \{(0,y)| y \in\mathbb{R}\}$ and $V_2'=\{(x,x)|x \in \mathbb{R}\}$.