Let $\mathbb{F}[X]$ be the polynomial ring with one variable.
$V$ can be regarded as an $\mathbb{F}[X]$-module by defining $Xv = T(v)$ for every $v \in V$. $\mathbb{F}[X]$-submodules of $V$ are none other than $T$-invariant subspaces of $V$.
Let $K = \mathbb{F}[X]/(p_t)$.
Since $p_t$ is irreducible, $K$ is a field.
Since $p_t V = 0$, $V$ can be regarded as a $K$-module.
Let $W$ be a $T$-invariant subspace of $V$.
$W$ can be regarded as a $\mathbb{F}[X]$-submodule of $V$.
Since $p_t W = 0$, $W$ can be regarded as a $K$-submodule of $V$.
Hence there exists a $K$-submodule $W'$ such that $V = W \oplus W'$.
Since $W'$ is a $\mathbb{F}[X]$-submodule, it is $T$-invariant.
This completes the proof.
For your first question, assume that the characteristic polynomial had an irreducible factor $p_i$ that does not divide the minimal polynomial. Then extending the ground field (by writing down a matrix for the endomorphism and interpreting it as one of a linear transformation $\phi$ over the larger field) we can obtain a root of $p_i$, which would be an eigenvalue of $\phi$ (a root of its characteristic polynomial, which hasn't changed by the field extension) without being a root of its minimal polynomial (which hasn't changed either). But every eigenvalue of $\phi$ must be a root of its minimal polynomial, since any polynomial $P(\phi)$ of $\phi$ acts on an eigenvector $v$ with eigenvalue $\lambda$ by the scalar $P(\lambda)$, and when $P$ is the minimal polynomial one has $P(\phi)=0$, whence $P(\lambda)=0$.
For your second question, $p_i$ divides every $q_j$ with $j\neq i$, so if $a_i=0$ then one would have that $p_i$ divides $a_1 q_1 + \dotsb + a_k q_k = 1$, which is absurd. (You don't need to apply wlog to get $a_i\neq0$.)
In reply to the added question, here is a proof for the first question entirely working over the original field. It will be a hardly inspiring induction on the dimension, where in addition it is hard to see where exactly is the crux of the proof, but that is because of a real difficulty: what exactly can we say about the characteristic polynomial $\chi_\phi$ of $\phi\in\operatorname{End}(V)$ without mentioning its roots, the eigenvalues (which may not exist over the original field)? I will use the following fact, which follows from the computation of the characteristic polynomial of a block triangular matrix: if $W$ is a $\phi$-stable subspace of $V$, then $\chi_\phi$ is the product of the characteristic polynomials of the endomorphisms $\phi|_W$ of $W$ and $\phi_{V/W}$ of $V/W$ defined by $\phi$. In this situation it is also clear that if some polynomial of $\phi$ vanishes (on $V$) then the same polynomial of $\phi|_W$ and of $\phi_{V/W}$ also vanish (on $W$ and $V/W$ respectively; the converse may fail), so the minimal polynomials of $\phi|_W$ and of $\phi_{V/W}$ divide that of $\phi$.
Now I will prove "every irreducible factor $p_i$ of $\chi_\phi$ divides the minimal polynomial of $\phi$" by induction on $\dim V$. If $\dim V=0$ then both polynomials are unity and there is nothing to prove. Otherwise choose a nonzero vector $w$ and let $W$ be the span of all vectors $\phi^i(w)$ for $i\geq0$ (the smallest $\phi$-stable subspace containing $w$). If $d$ is minimal such that $\phi^d(w)$ is linearly dependent on $w,\phi(w),\ldots,\phi^{d-1}(w)$ then $d=\dim W$ is nonzero since $w\neq0$, and the minimal polynomial $P$ of $\phi|_W$ cannot be of degree less than $d$; therefore it is of degree $d$ and equal to the characteristic polynomial of $\phi|_W$. Since $P$ divides the minimal polynomial of $\phi$, we are done if $p_i$ divides $P$, so assume this is not the case. But then $p_i$ divides the characterisitic polynomial of $\phi_{V/W}$; by the induction hypothesis it divides the minimal polynomial of $\phi_{V/W}$, and therefore the one of $\phi$.
Although I appeared to use the Cayley-Hamilton theorem in concluding equality of minimal and characteristic polynomials of $\phi|_W$, this can in fact be proved by a direct calculation (the matrix in the basis $w,\phi(w),\ldots,\phi^{d-1}(w)$ is a companion matrix), and a proof of the Cayley-Hamilton theorem along these lines is given in this answer.
Best Answer
This is very simple. Suppose $W$ is an invariant subspace for$~\alpha$, then one can restrict $\alpha$ to$~W$ to obtain an endomorphism $\alpha|_W$ of$~W$. Then the characteristic polynomial of $\alpha|_W$ divides the characteristic polynomial of $\alpha$, contradicting (if $0\subset W\subset V$) the assumption that the latter is irreducible.
The fact that the characteristic polynomial of the restriction to an invariant subspace$~W$ divides that of the full endomorphism is immediate from the definition of the characteristic polynomial, using a basis of $V$ that extends a basis of$~W$ to get a matrix for the endomorphism: this matrix is block upper triangular, with a matrix for the restricted endomorphism as upper left diagonal block.