= Solution
Use the <pair of partitions for a generalized Specht module> convention appropriate to <generalized Specht modules>. Put $a=\mu^\sharp$ and $b=\mu$. Here $b=(b_1,b_2,\ldots)$ is a sequence of nonnegative row lengths of total $n$, and $a$ is a proper <partition of an integer> satisfying $0\le a_i\le b_i$ and $a_{i+1}\le a_i$. The row-length sequence $b$ is allowed to be a composition; it need not decrease. The first $a_i$ cells of row $i$ are marked. This construction is not an arbitrary pair of unrelated <partitions of an integer> and is not simply a skew diagram $b/a$.
A word has type $b$ if it contains $b_i$ occurrences of $i$. Read it from left to right. Every occurrence of $1$ is good; an occurrence of $i+1$ is good exactly when there have so far been more good occurrences of $i$ than good occurrences of $i+1$. Otherwise it is bad. Define
$$
s(a,b)=\{w:\operatorname{type}(w)=b,\ \#\text{good occurrences of }i\ge a_i\text{ for every }i\}.
$$
Thus $s(0,b)$ is the set of all words of type $b$, and $s(b,b)$ consists of <lattice words> when $b$ is a proper <partition of an integer>. Marking the whole first row makes no difference because every $1$ is good.
For a <Young tableau> $T$ with row lengths $b$, let $C_T^a$ permute labels within each column of its marked subdiagram, fixing all unmarked labels. Its generalized <polytabloid> and <generalized Specht module> are
$$
\boxed{e_T^{a,b}=\sum_{g\in C_T^a}\operatorname{sgn}(g)\{gT\},\qquad
S^{a,b}=\operatorname{span}_F\{e_T^{a,b}\}\subseteq M^b.}
$$
In particular $S^{0,b}=M^b$ and $S^{b,b}=S^b$. The <Young tableau> shape in this definition must be the row sequence $b$.
We give the filtration argument, keeping its combinatorial and module-theoretic steps separate. Normalize $a_1=b_1$, and if $a\ne b$ choose the first row $c>1$ with $a_c<b_c$. Then $a_{c-1}=b_{c-1}$. Define $A_c(a,b)=(a+e_c,b)$ if $a_{c-1}>a_c$; otherwise the add branch is empty. Define $R_c(a,b)=(a,b')$ by moving the unmarked tail of row $c$ into row $c-1$:
$$
b'_c=a_c,\qquad b'_{c-1}=b_{c-1}+b_c-a_c,
$$
with other row lengths unchanged. Re-normalize the first marked row if necessary. The Basic Combinatorial Theorem supplies the counting and formal-character recursions
$$
|s(a,b)|=|s(A_c(a,b))|+|s(R_c(a,b))|,\qquad
[0]^{[a,b]}=[0]^{[A_c(a,b)]}+[0]^{[R_c(a,b)]}.
$$
At a terminal pair $(\nu,\nu)$ the formal term is $[\nu]$. Thus the second recursion specifies the Specht multiplicities by the terminal leaves, with repeated leaves counted separately.
Define an equivariant map $\psi:M^b\to M^{b'}$ by summing, on each <tabloid>, over all choices of the $a_c$ labels retained in row $c$, moving the remaining labels to row $c-1$. Column cancellation gives
$$
\psi(S^{a,b})=S^{R_c(a,b)},\qquad
S^{A_c(a,b)}\subseteq S^{a,b}\cap\ker\psi.
$$
For the first identity, terms that put two marked labels of one column in the raised row cancel in pairs; the surviving marked antisymmetrizer is the one for the raised pair, and every target generator has a preimage. For the second, the extra marked cell makes such a collision unavoidable, so the sum vanishes. Also antisymmetrizing the extra cell expresses its generator as a sum of old generators, proving containment in $S^{a,b}$.
There is a characteristic-independent lower bound $\dim S^{a,b}\ge|s(a,b)|$. Given a word in $s(a,b)$, place the label $j$ in its letter's row: good occurrences fill that row from the left and bad occurrences fill it from the right. The marked cells form increasing column chains because each good $i+1$ has an earlier unmatched good $i$. Its generalized <polytabloid> has the original row-assignment <tabloid> with coefficient one; every other term changes an earlier member of a marked chain. Ordering these row assignments gives a triangular coefficient matrix. The resulting vectors are independent over every <field>.
Start with $(0,v)$, where equality holds because the <tabloid> basis is indexed by all words of type $v$. Any pair is reached from such a pair by a sequence of add and raise operations: reverse a raise by splitting a raised tail, and reverse an add by unmarking a cell; row by row these recover an entirely unmarked composition. Suppose equality holds at a pair. The map and <kernel> containment above give
$$
|s(a,b)|=\dim S^{a,b}\ge\dim S^{A_c(a,b)}+\dim S^{R_c(a,b)}
\ge|s(A_c(a,b))|+|s(R_c(a,b))|=|s(a,b)|.
$$
Every inequality is therefore equality. In particular the containment is the whole <kernel>, and
$$
0\longrightarrow S^{A_c(a,b)}\longrightarrow S^{a,b}\longrightarrow S^{R_c(a,b)}\longrightarrow0
$$
is exact. The process terminates because each add marks an extra cell and each raise decreases the total row index of unmarked cells. Induction from the terminal <Specht modules>, splicing their filtrations through these exact sequences, proves \b[a <Specht series> with factors precisely $[0]^{[\mu^\sharp,\mu]}$]. No splitting of these sequences in positive <characteristic> is asserted.
Applying the same marking, raising and straightening procedure to the two sets of <column antisymmetrizers> in $(S^\mu\boxtimes S^\lambda)\uparrow^{S_{r+s}}$, where $|\mu|=r$, $|\lambda|=s$, leaves the cells of $\mu$ fixed and records the added cells by their row labels from $\lambda$. The column relations require strict increase down columns, the row relations weak increase along rows, and the <good-letter matching in a tableau word> test requires a <lattice word>. Thus the surviving terminal multiplicity is the number $c_{\mu\lambda}^\nu$ of <Semistandard Young tableaux> of <skew shape> of shape $\nu/\mu$ and content $\lambda$ whose word, read right-to-left along successive rows from top to bottom, has at least as many $i$'s as $(i+1)$'s in every prefix. The same triangular leading-tabloid argument identifies each quotient with $S^\nu$. Consequently the <Littlewood–Richardson rule> is
$$
\boxed{[\mu][\lambda]=\sum_{\nu\vdash r+s}c_{\mu\lambda}^\nu[\nu].}
$$
It describes ordinary <irreducible> constituents in <characteristic> zero and <Specht filtration> factors over arbitrary $F$.
For induction by one letter, content $(1)$ permits exactly one added cell and each multiplicity is one. A <Specht series> for the requested induced <module> has successive quotients, ordered by addable nodes from bottom to top,
$$
\boxed{S^{(4,2,2,1,1)},\quad S^{(4,2,2,2)},\quad S^{(4,3,2,1)},\quad S^{(5,2,2,1)}.}
$$
Equivalently there is $0=N_0\subset N_1\subset N_2\subset N_3\subset N_4=S^{(4,2,2,1)}\uparrow^{S_{10}}$ with these four quotients in order. Their <dimensions> $567,300,768,525$ sum to $2160=10\cdot216$, the <dimension> of the induced <module>.
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