You should start realising how little you are asked to do: it is given that $c$ is an eigenvalue (characteristic value), so a root of$~p_T$, and the (monic) factor $q$ can absorb anything; therefore all that is asked is to show that $c$ is a multiple root of the minimal polynomial $p_T$. Although I initially thought that "characteristic subspace" meant "generalized eigenspace" (blame my French connection, since that is precisely what "sous espace caractéristique" means), I will take it that $W_c$ here is just the ordinary eigenspace for$~c$ of$~T$, as with the other interpretation the hypothesis of question becomes absurd as we shall see.
It is given that for some $T$-stable subspace $W$ containing the characteristic subspace$~W_c$, there exists $\alpha\in V\setminus W$ with $(T-cI)v\in W$. This means that the image of $\alpha$ in the quotient space $V/W$ is an eigenvector with eienvalue$~c$ for the linear map $T_{/W}$ induced by $T$ in $V/W$. The existence of such a vector shows that the algebraic multiplicity of$~c$ as eigenvalue of$~T$ (on all of $V$) is strictly larger than its algebraic multiplicity as eigenvalue of the restriction$~T|_W$ of$~T$ to$~W$. This is because the characteristic polynomial of$~T$ is the product of the characteristic polynomials of $T|_W$ and of $T_{/W}$, the latter having$~c$ as root. But since $W$ already contains the whole eigenspace $W_c$, the geometric multiplicities of$~c$ as eigenvalue of $T|_W$ and of $T$ are the same. Hence the generalised eigenspace of$~T$ cannot be$~W_c$ (and in particular $T$ cannot be diagonalisable), nor indeed can it be contained in$~W$. Then there is some vector $v\in V$ with $(T-cI)^2(v)=0$ but $(T-cI)(v)\neq0$, which shows that $c$ is indeed a multiple root of$~p_T$.