= Solution
By part (a), work first with an affine base and its homogeneous equations. Put $S=k[y_0,\ldots,y_n]$, write $S_N$ for its finite-dimensional degree-$N$ <vector space>, and use $S_j=0$ for $j<0$. At each base point define the <graded multiplication map for a projective fiber>
$$
\mu_N(x):\bigoplus_{i=1}^r S_{N-d_i}\longrightarrow S_N,\qquad (g_i)_i\longmapsto\sum_i g_i f_i(x,y).
$$
Its image is exactly $(I_x)_N$, since taking the degree-$N$ part of any expression in the <homogeneous ideal> discards all other degrees. Moreover
$$
\mathfrak m^N\subseteq I_x\iff(I_x)_N=S_N\iff\mu_N(x)\text{ is surjective},
$$
because degree-$N$ <monomials> generate $\mathfrak m^N$.
Choose <monomial bases> for the domain summands and codomain. Every entry of the resulting <matrix> is a coefficient of one of the $f_i$, hence a <regular function> of $x$. If $h_N=\dim S_N$, failure of surjectivity is the vanishing of all $h_N$-by-$h_N$ <matrix minors>. It is therefore a closed rank-defect locus $C_N$. If the <matrix> has fewer than $h_N$ columns, the failure is automatic and $C_N=X$, also closed. Constants and $N=0$ are covered by the same construction.
Part (c) gives $\pi(Y)=\bigcap_{N\ge0}C_N$, an intersection of closed subsets, hence closed. Apply part (a) to pass back to arbitrary $X$. Thus \b[projection from $X\times\mathbb P^n$ sends every closed subset to a closed subset]. This proves the <closedness of projection from projective space> without assuming the properness result being proved.
The field convention matters: a <rational-point projection need not be Zariski closed>. Over $\mathbb R$, the closed equation $y_0^2=t y_1^2$ in $\mathbb A^1\times\mathbb P^1$ has real-point image $[0,\infty)$. Indeed $y_1=0$ would force $y_0=0$, so a solution exists precisely when $t$ is a square in $\mathbb R$. This infinite proper subset of the affine line is not <Zariski closed>. The valid arbitrary-field theorem uses scheme or geometric images, for which the matrix argument remains valid at every residue field.
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