$\dim\ker T = \dim\ker T^{\vee}$ 

Let $V,W$ be vector spaces over $\mathbb{F}$, and $T:V\to W$ a linear map between the spaces. Let $T^{\vee}:W^{\vee}\to V^{\vee}$ denote the dual map of $T$ ($V^{\vee}, W^{\vee}$ are the dual space of $V,W$ respectively).
    True/false: $\dim\ker T = \dim\ker T^{\vee}$


I say the claim is true. Proof:
Let $\mathcal{B} = (v_1,\ldots,v_n)$ be a basis for $V$, and $\mathcal{C}=(w_1,\ldots,w_k)$ a basis for $W$, and let $A$ be the matrix representation of $T$ relative to $\mathcal{B},\mathcal{C}$. Let $\mathcal{B}^{\vee},\mathcal{C}^{\vee}$ be the dual bases of $\mathcal{B},\mathcal{C}$. Then the matrix representation of $T^{\vee}$ w.r.t $\mathcal{C}^{\vee},\mathcal{B}^{\vee}$ is $A^t$. Since $rank(A) = rank(A^t)$, from the rank-nullity theorem
$$
n = rank(A) + \dim\ker{A} = rank(A^t) + \dim\ker{A^t} = rank(A) + \dim\ker{A^t}
$$
And therefore, $\dim\ker A = \dim\ker A^t$, and therefore $\dim\ker T = \dim\ker T^{\vee}$.
Is this proof OK? Does this also imply that $\dim ImT = \dim ImT^{\vee}$?

Yes the proof is correct. The crucial point is to show that the matrix associate to $\text{}^tT$ is $\text{}^tA$ where $A$ is the matrix associate to $T$.
