= Solution
We prove <fourth-moment tightness of polygonal random walks> uniformly, including intervals shorter than a mesh cell or crossing a mesh boundary. For $s<t$, let $c_j$ be the length of the overlap of $(Ns,Nt)$ with $(j-1,j)$. The interpolated increment is
$$
S_t^N-S_s^N=N^{-1/2}\sum_jc_j\xi_j,
\qquad 0\le c_j\le1,\qquad \sum_jc_j=N(t-s).
$$
Put $m_4=\mathbb E\xi^4<\infty$. Expanding the fourth power and using centered <independent> steps, only the terms with four equal indices or two equal pairs survive. Consequently
$$
\mathbb E|S_t^N-S_s^N|^4
=N^{-2}\left(m_4\sum_jc_j^4+6\sum_{i<j}c_i^2c_j^2\right)
=N^{-2}\left(3\left(\sum_jc_j^2\right)^2+(m_4-3)\sum_jc_j^4\right).
$$
If $m_4\le3$, the last correction is nonpositive. If $m_4>3$, use $\sum c_j^4\le(\sum c_j^2)^2$. In either case the expression is at most $\max(3,m_4)N^{-2}(\sum c_j^2)^2$. Since $c_j^2\le c_j$,
$$
\boxed{\mathbb E|S_t^N-S_s^N|^4\le\max(3,m_4)|t-s|^2,\quad\text{for all }N,s,t.}
$$
Also $S_0^N=0$ and the paths are continuous. The stated tightness criterion applies with $\gamma=4$, $\alpha=1$ and the displayed constant, so the laws $\mu_N$ are tight on $C_0([0,1],\mathbb R)$ with its uniform topology. Part (a) established all the Brownian <finite-dimensional distribution> limits. The supplied implication from tightness plus those limits now yields the <Donsker invariance principle>:
$$
\boxed{\mu_N\Rightarrow\mu\quad\text{in }C([0,1],\mathbb R),}
$$
where $\mu$ is <Wiener measure>. No estimate restricted to grid endpoints would suffice by itself for the printed uniform criterion; the overlap coefficients handle every pair of times.
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