Elementary property preserved under vector space isomorphism 
Suppose $T:V\to W$ be an isomorphism. Prove:
$(1)$ $S\subseteq V$ is independent$\iff$$T(S)\subseteq W$ is independent.
$(2)$ $U\leq V$ and $\mathrm{dim}(U)=n\in \Bbb{N}$$\iff$$T(U)\leq W$ and $\mathrm{dim}(T(U))=n\in \Bbb{N}$.
$(3)$ Let $S\subseteq V$. Then $\alpha \in \mathrm{span}(S)$$\iff$$T(\alpha)\in \mathrm{span}(T(S))$.

My attempt: $(1)$ Suppose $S\subseteq V$ is independent. Since $T$ is injective, $T(S)$ is independent (here is proof). Conversely, suppose $T(S)\subseteq W$ is independent. Since $T$ is linear bijective, we have $T^{-1}:W\to V$ is linear bijective, by theorem 7 section 3.2. In particular, $T^{-1}$ is injective. So $T^{-1}(T(S))$ $=T^{-1}\circ T(S)$ $=\text{id}_V(S)=S$ is independent.
$(2)$ Suppose $U\leq V$ and $\mathrm{dim}(U)=n\in \Bbb{N}$. Here is definition of vector subspace. Since $U\neq \emptyset$, we have $T(U)\neq \emptyset$. Let $c\in F$ and $u,v\in T(U)$. Then $\exists x,y\in U$ such that $T(x)=u$, $T(y)=v$. So $c\cdot u+v$ $=c\cdot T(x)+T(y)$. Since $T$ is linear, we have $c\cdot T(x)+T(y)$ $=T(c\cdot x+y)$. Since $U\leq V$, we have $c\cdot x+y\in U$. Thus $c\cdot u+v=T(c\cdot x+y)\in T(U)$. Hence $T(U)\leq W$. Now we show $\mathrm{dim}(U)= \mathrm{dim}(T(U))$. Approach(1): let $\{\alpha_1,…,\alpha_n\}$ be basis of $U$. We claim $\{T(\alpha_1),…,T(\alpha_n)\}$ is basis of $T(U)$. Let $y\in T(U)$. Then $\exists x\in U$ such that $T(x)=y$. Since $\mathrm{span}(\{\alpha_1,…,\alpha_n\})=U$, we have $x=\sum_{i\in J_n}a_i\cdot_V \alpha_i$. Since $T$ is linear, we have $T(x)$ $=T(\sum_{i\in J_n}a_i\cdot_V \alpha_i)$ $= \sum_{i\in J_n}a_i\cdot_W T(\alpha_i)$ $=y$. Thus $y\in \mathrm{span}(\{T(\alpha_1),…,T(\alpha_n\})$. Hence $T(U)= \mathrm{span}(\{T(\alpha_1),…,T(\alpha_n\})$. Since $\{\alpha_1,…,\alpha_n\}$ is independent and $T$ is injective, we have $T(\{\alpha_1,…,\alpha_n)\}$ $=\{T(\alpha_1),…,T(\alpha_n)\}$ is independent. Thus $\{T(\alpha_1),…,T(\alpha_n)\}$ is basis of $T(U)$. Hence $\mathrm{dim}(U)$ $= \mathrm{dim}(T(U))$ $=n$. Approach(2): Define $h:U\to T(U)$ such that $h(x)=T(x)$, $\forall x\in U$. It’s easy to check, $h$ is linear bijective. So $h$ is isomorphism. $U$ is finite dimensional vector space. By exercise 6 section 3.3, $\mathrm{dim}(U)$ $= \mathrm{dim}(T(U))$ $=n$.
Conversely, suppose $T(U)\leq W$ and $\mathrm{dim}(T(U))=n\in \Bbb{N}$. Since $T(U)\neq \emptyset$, we have $T^{-1}(T(U))=U\neq \emptyset$. Let $c\in F$ and $u,v\in U$. Since $T$ is invertible, we have $\exists T(u),T(v)\in T(U)$ such that $T^{-1}(T(u))=u$ and $T^{-1}(T(v))=v$. So $c\cdot u+v$ $=c\cdot T^{-1}(T(u))+ T^{-1}(T(v))$. By theorem 7 section 3.2, $T^{-1}$ is a linear map. So $c\cdot T^{-1}(T(u))+ T^{-1}(T(v))$ $=T^{-1}(c\cdot T(u)+T(v))$. Since $T(U)\leq W$, we have $c\cdot T(u)+T(v)\in T(U)$. Thus $c\cdot u+v$ $=T^{-1}(c\cdot T(u)+T(v))\in$ $T^{-1}(T(U))=U$. Hence $U\leq V$.Define $h:T(U)\to U$ such that $h(y)=T^{-1}(y)$, $\forall y\in T(U)$. It’s easy to check, $h$ is linear bijective. So $h$ is isomorphism. $T(U)$ is finite dimensional vector space. By exercise 6 section 3.3, $\mathrm{dim}(T(U))$ $= \mathrm{dim}(U)=n$.
$(3)$ Suppose $\alpha\in \mathrm{span}(S)$. Then $\alpha =\sum_{i\in J_n}a_i\cdot_V u_i$, where $n\in \Bbb{N}$, $a_i\in F$, $u_i\in S$. Since $T$ is a linear map, we have $T(\alpha)$ $=T(\sum_{i\in J_n}a_i\cdot_V u_i)$ $=\sum_{i\in J_n}a_i\cdot_W T(u_i)$. Since $u_i\in S$, we have $T(u_i)\in T(S)$. So $T(\alpha)$ $=\sum_{i\in J_n}a_i\cdot_W T(u_i)$ $\in \mathrm{span}(T(S))$. Conversely, suppose $T(\alpha)\in \mathrm{span}(T(S))$. Then $T(\alpha)$ $=\sum_{i\in J_n}a_i\cdot_W u_i$, where $n\in \Bbb{N}$, $a_i\in F$, $u_i\in T(S)$. Since $u_i\in T(S)$, we have $\exists x_i\in S$ such that $T(x_i)=u_i$, $\forall i\in J_n$. So $T(\alpha)$ $=\sum_{i\in J_n}a_i\cdot_W u_i$ $= \sum_{i\in J_n}a_i\cdot_W T(x_i)$ $=T(\sum_{i\in J_n}a_i\cdot_V x_i)$. Since $T$ is injective, we have $\alpha = \sum_{i\in J_n}a_i\cdot_V x_i$. Thus $\alpha \in \mathrm{span}(S)$. Is my proofs correct?
Do you known any other property to keep in arsenal?
 A: Isomorphism : isos "equal", and "morphe "structure" (see here)
Here mathematical structure is  vector space (linear space).
An isomorphism $T:V_F\to W_F$ is a bijective linear map.
Bijective map preserve set theoretic structure . $V, W$ are isomorphic ( have same cardinality) merely as sets.
The linear map and it's inverse (which is also linear) preserve the linear structure of the vector space. In other words of two vector spaces are isomorphic then both have the same linear structure. Hence all the properties that depends on the linearity are preserved.
For an example :

*

*$S\subset V$ LI iff $T(S) \subset W$ is LI.


*$S$ spans $V$ iff $T(S) $ spans $W$ .


*$B$ is basis of $V$ iff $T(B) $ is basis of $W$


*$\dim V=\dim W $
etc.

All your proofs are correct.
Summary of your proofs:


*$T:V\to W$ isomorphism implies $T|_U:U\to T(U) $ is also an isomorphism.

Hence
$U\cong T(U) $


*$\text{span}({S})$ is also a subspace of $V$ .

Hence $\text{span}({S})\cong T(\text{span}(S))=\text{span}(T(S))$
