= Solution
If $c(\lambda)=\det(\lambda I-T)=0$, then $\lambda I-T$ has nonzero <kernel of a linear map>. A nonzero vector $v$ in that <kernel of a linear map> satisfies $Tv=\lambda v$, proving that $\lambda$ is an <eigenvalue>. Extend $v$ to a <basis>. Its first column is $(\lambda,0,\ldots,0)^T$, so the <matrix> has block form
$$
[T]=\begin{pmatrix}\lambda&*\\0&B\end{pmatrix},\qquad c(X)=(X-\lambda)\det(XI-B).
$$
Here $B$ represents the induced operator on $V/\langle v\rangle$.
If the <algebraic multiplicity> is $r$, repeat this argument in that quotient: its <characteristic polynomial> still has $\lambda$ with multiplicity $r-1$. Induction and lifting a quotient <basis> gives a <basis> whose first $r$ columns have the prescribed upper-triangular form. With $U$ the span of those $r$ vectors, the full <matrix> is
$$
\begin{pmatrix}\lambda I+N&*\\0&C\end{pmatrix},\qquad N^r=0,\qquad\det(C-\lambda I)\ne0.
$$
Thus $U$ is invariant and is contained in $W_\lambda$. Conversely, if $(T-\lambda I)^sv=0$, project to $V/U$. The induced map $C-\lambda I$ is invertible, so its $s$th power cannot kill a nonzero class. Hence $v\in U$. Consequently $\boxed{W_\lambda=U,\ \dim W_\lambda=r}$; this proves simultaneously that it is a <vector subspace>[subspace] and $T$-invariant. It also proves $(T-\lambda I)^r=0$ on this space, without using <Cayley-Hamilton theorem> in advance.
To prove independence of the different spaces, take a relation $\sum_\lambda v_\lambda=0$ with $v_\lambda\in W_\lambda$. Fix $\lambda$ and apply $P_\lambda(T)=\prod_{\mu\ne\lambda}(T-\mu I)^{r_\mu}$. It kills all terms except $v_\lambda$. On $W_\lambda$, each factor $T-\mu I=(\lambda-\mu)I+N$ is invertible: its inverse is a finite <geometric series> in the <nilpotent operator>[nilpotent] $N$. Hence $P_\lambda(T)v_\lambda=0$ implies $v_\lambda=0$. The sum is direct, and its <dimension> is $\sum r_\lambda=\dim V$, so
$$
\boxed{V=\bigoplus_\lambda W_\lambda.}
$$
Finally $c(T)=\prod_\lambda(T-\lambda I)^{r_\lambda}$ kills each summand, hence every vector. This proves $\boxed{c(T)=0}$ without circularly assuming the theorem during the decomposition.
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