Proof Verification: Prove that a space is finite dimensional. 

If $T$ is a linear map from $V$ to $W$ such that $\ker(T)$ and $\operatorname{im}(T)$ are both finite dimensional. Then $V$ is also finite dimensional.


Since the proof of dimension theorem requires $V$ to be finite dimensional, we cannot directly use it here.
Sketch of my proof: Since I havent been introduced with other tools about infinite dimensional spaces, I rely on the fact that if $V$ is infinite dimensional, then there exists an infinite sequence of vectors $\vec v_1,\vec v_2,\dots \in V$ such that $(\vec v_1,\dots ,\vec v_n)$ is linearly independent for any positive integer $n$.
First I construct a basis $(\vec u_1,\dots ,\vec u_m)$ for $\ker(T)$, then proceed to prove that all the lists in the form: $(\vec u_1,\dots, \vec u_m, \vec v_1,\dots , \vec v_n)$ is linearly independent if $\vec v_j $ are not in $\ker(T)$.
Then we can prove that $(T\vec v_1,\dots , T\vec v_n)$ is linearly independent for any value of $n$. Which is then equivalent to $\operatorname{im}(T)$ being infinite dimensional. So we have arrived at a contradiction.
Also, the question looks so obvious, is there a way to show this with a more clever method?
 A: Here is a quick sketch of a proof, leaving out all the actual linear algebra details and concentrating on the construction of the appropriate bases.
Choose a basis $\{a_i\}_{i \in I}$ for $\text{kernel}(T)$. 
Choose a basis $\{b_j\}_{j \in J}$ for $\text{image}(T)$ (make sure your index sets $I,J$ are disjoint). 
Choose $b'_j \in T^{-1}(b_j)$, for each $j \in J$. 
Note that $a_i \ne b'_j$ for all $i \in I$ and $j \in J$, because $a_i \in \text{kernel}(T)$ but $b_j \ne \vec 0$ and therefore $b'_j \not\in \text{kernel}(T)$.
Finally, $\{a_i\}_{i \in I} \cup \{b'_j\}_{j \in J}$ is a basis for $V$ (This is where all the linear algebra details are hidden).
So far there is no assumption on finite dimension.
But if the kernel is finite dimensional then $I$ is a finite set, and if the image is also finite dimensional then $J$ is a finite set, and therefore $V$ has a finite basis indexed by the finite set $I \cup J$. 
By the way, one can also use this argument to construct an isomorphism between $V$ and $\text{kernel}(T) \oplus \text{image}(T)$, taking the basis $\{a_i\}_{i \in I} \cup \{b'_j\}_{j \in J}$ bijectively to the basis $\{a_i\}_{i \in I} \cup \{b_j\}_{j \in J}$. This works without any assumption on cardinality of bases.
