I think this will work:
Suppose $V$ is a vector space over the field $\Bbb F$, with $\dim V = N < \infty$. Then the minimal polynomial of $T$ is the monic polynomial $m(x) \in \Bbb F[x]$ of least degree such that $m(T) = 0$ identically on $V$. As such, we have $m(T) = 0$ on $W$ as well, whence $m(T_{\vert W}) = 0$. Now let $m_W(x) \in \Bbb F[x]$ be the minimal polynomial of the restriction $T_{\vert W}$ of $T$ to $W$. Since $\Bbb F$ is a field, the usual division algorithm for polynomials holds in $\Bbb F[x]$. Thus we may write $m(x) = m_W(x)q(x) + r(x)$ for some unique $q(x), r(x) \in \Bbb F[x]$ with either $r(x) = 0$ or $0 \le \deg r(x) < \deg m_W(x)$, whence $r(x) = m(x) - m_W(x)q(x)$. Then $r(T_{\vert W}) = m(T_{\vert W}) - m_W(T_{\vert W})q(T_{\vert W}) = 0$. Now in the event $r(x) \ne 0$, let the leading coefficient of $r(x)$ be $\beta \in \Bbb F$. Set $r'(x) = \beta^{-1} r(x)$. Then $r'(x)$ is monic, $\deg r'(x) = \deg r(x)$, and furthermore $r'(T_{\vert W}) = \beta^{-1} r(T_{\vert W}) = 0$. But since $\deg r'(x) < \deg m_W(x)$, this contradicts the minimality of $m_W(x)$ unless $r'(x) = \beta^{-1}r(x) = 0$. Thus $r(x) = 0$ and hence $m_W(x) \mid m(x)$. QED
Note Added in Edit, Sunday 29 August 2021 10:16 PM PST: Though the above proof seems pretty straightforward to me, by and large, I feel the assertion made in the third sentence, that
"As such, we have $m(T) = 0$ on $W$ as well, whence $m(T_{\vert W}) = 0$,"
may be further clarified. We observe that the given $T$-invarinace of
$W \subset V \tag{1}$
allows us to write
$Tw \in W \tag 2$
for any
$w \in W; \tag 3$
also, for any such $w$, by definition we have
$T_{\vert W} w = Tw, \tag 4$
and thus by virtue of (2) we may write
$T_{\vert W}Tw = TTw = T^2w, \tag{4.5}$
and hence, in accord with (2)-(4.5), we further have
$T^2_{\vert W} w = T_{\vert W} T_{\vert W} w = T_{\vert W} Tw = TTw = T^2w; \tag 5$
indeed, at this point we may allow a simple induction to take over, assuming
$T^k_{\vert W} w = T^k w \tag{6}$
for some $k \in \Bbb N$ and any $w \in W$; then since $T^k_{\vert W} w \in W$ we may write
$T^{k + 1}_{\vert W} w = TT^k_{\vert W} w = TT^k w = T^{k + 1}w; \tag 7$
from this it is easy to see that for any
$p(x) \in \Bbb F[x] \tag 8$
$p(T_{\vert W})w = p(T)w \tag 9$
provided $w \in W$; and from this we readily conclude that
$m(T)w = m(T_{\vert W})w = 0 \tag{10}$
as well. End of Note.
Cheers, and as always,
Fiat Lux!!!
Best Answer
We begin by noting that for vector spaces $X,Y$, a subspace $S \subset X$ and a linear map $F \colon X \to Y$, we have an induced map
$$\tilde{F} \colon X/S \to Y$$
that satisfies $\tilde{F}(x+S) = F(x)$ for all $x\in X$ if and only if $S \subset \ker F$. [prove it, or cite a theorem]
Then we apply the above to the situation $S = U$, $X = V$, and $Y = V/U$, with $F = \pi\circ T$, where $\pi \colon V \to V/U$ is the canonical projection.
For the remaining part, note that $\widetilde{\pi\circ p(T)} = p(\widetilde{\pi\circ T}) = p(B)$ for all polynomials $p\in K[X]$. [prove it, or cite a theorem]