If $\dim(V) $ is infinite, show that $V\oplus V$ is isomorphic to $V$

Your argument is good, you just need to finish it up.

Assuming choice, you can consider a basis $\mathcal{B}$ of $V$. Then the set $\mathcal{C}=\{(v,0):v\in\mathcal{B}\}\cup\{(0,v):v\in\mathcal{B}\}$ is a basis for $V\oplus V$.

Since $\mathcal{B}$ is infinite, there is a bijection $f\colon\mathcal{C}\to\mathcal{B}$, because $\mathcal{C}$ is just the disjoint union of two copies of $\mathcal{B}$. Then the linear map $T\colon V\oplus V\to V$, defined on $\mathcal{C}$ by $T(w)=f(v)$ and extended by linearity, is an isomorphism, because it sends a basis onto a basis.

Note that choice is needed in two places: to obtain a basis for $V$ and to use that $|X|+|X|=|X|$ when $X$ is an infinite set.

Since your vector space $V$ is infinite-dimensional, let $b_1, b_2, \dots, b_n, \dots$ be a basis of $V$. Then the set $$(B \oplus 0) \cup (0 \oplus B) = \{b_1 \oplus 0, b_2 \oplus 0, \dots, 0 \oplus b_1, 0 \oplus b_2, \dots\}$$

forms a basis of of $V \oplus V$.

Assuming the dimension of $V$ is countable, the dimension of $V \oplus V$ is countable too. Can you come up with an invertible linear transformation yourself, by thinking of how you show that $|\mathbb{N}| = |\mathbb{Z}|$?