Formal solution without handwaving about Jordan normal form Let $A$ be a $7\times 7$ matrix over $\mathbb C$ with minimal polynomial $(t-2)^3$. I need to prove $\dim \ker (A-2)\geq 3$.
The handwavy argument I have is that $\deg m$ is the size of the greatest Jordan block while $\dim \ker (A-2)$ is the number of blocks, and since $2\cdot 3<7$, the dimension must be at least $3$. However, I realized I don't know how to formally prove this, i.e without taking the sentences I said as facts.

I wouldn’t call your argument handwavy, but we can replace the use of the  Jordan normal form by the underlying calculations:

Lemma: Let $V$ be a finite-dimensional vector space and $f, g \colon V \to V$ be two endomorphisms. Then $\dim \ker (fg) \leq \dim \ker f + \dim \ker g$.
Proof: We have $g( \ker(fg) ) \subseteq \ker f$ and thus by the dimensional formula that
  $$
        \dim \ker (fg)
  =     \dim \operatorname{im} g|_{\ker(fg)} + \dim \ker g|_{\ker(fg)}
  \leq  \dim \ker f + \dim \ker g.
$$

Thus we get the following by induction:

Corollary: If $V$ is a finite-dimensional vector space and $f_1, \dotsc, f_n \colon V \to V$ are endomorphisms, then
  $$
        \dim \ker(f_1 \dotsm f_n)
  \leq  \dim \ker f_1 + \dotsb + \dim \ker f_n.
$$
  We have in particular that $\dim \ker f^k \leq k \dim \ker f$ for every endomorphism $f \colon V \to V$ and every $k \in \mathbb{N}$.

If we apply this corollary to the endomorphism $f \colon \mathbb{C}^7 \to \mathbb{C}^7$, $x \mapsto (A-2) x$ we find that
$$
        7
  =     \dim \mathbb{C}^7
  =     \dim \ker f^3
  \leq  3 \dim \ker f
  =     3 \dim \ker (A-2),
$$
and thus $\dim \ker (A-2) \geq 3$.
