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The inclusion here is inclusion of Young diagrams, not the dominance order on partitions. The skew Young diagram $\lambda /\mu$ is the set difference of their cells. Two cells are adjacent when they share an edge. It is connected when any two cells can be joined by such steps, and is a rim hook when it is connected and contains no $2\times 2$ square. Under the usual edge-adjacency convention a totally disconnected skew Young diagram has only singleton components, equivalently no two cells share an edge. A horizontal strip instead means at most one cell per column; this distinction matters for the last part of this question.

A standard skew Young tableau is a linear extension of a partially ordered set: the cells are ordered by the row and column inequalities, and the tableau lists them in increasing label order. Write the current list as $x_1,\dots ,x_k$ and let $r_j$ be the desired label of $x_j$ in $R$. Then $(r_1,\dots ,r_k)$ is the one-line notation of the unique permutation $\pi$ with $\pi T=R$.

Whenever this list of desired labels is not increasing, there is an adjacent descent $r_j>r_{j+1}$. The cells $x_j,x_{j+1}$ are incomparable in the cell partial order: if they were comparable, both $T$ and $R$ would have to put them in the same order. Interchanging their consecutive current labels is therefore admissible. It removes exactly one inversion of a permutation from the list of desired labels. Repeating ends at $R$ after exactly the original inversion count, which is the Coxeter length $\ell (\pi )$. ThusThis proves the reduced adjacent-swap path between linear extensions and applies equally to ordinary tableaux.

With the usual meaning of totally disconnected skew Young diagram, the assertion in the PDF is false. For example, $\lambda =(3)$ and $\mu =(1)$ give two adjacent cells in one row, so the skew shape is connected, but its sole standard skew Young tableau affords the trivial representation of $S_2$. The multiplicity is one, not zero. The correct criterion is a horizontal strip:If the course uses “totally disconnected” specifically for no repeated column, its terminology must be read as this criterion; it cannot mean disconnected into individual cells.

For necessity, a non-horizontal strip has two vertically adjacent cells. They form a cover in the cell partial order, and some linear extension of a partially ordered set places them consecutively. To justify that assertion, put all predecessors of the lower cell other than the upper cell first, then the upper and lower cells, and complete the order; the cover property prevents any missing intermediate predecessor. The corresponding standard skew Young tableau has consecutive entries in one column. The relevant adjacent transposition acts on its vector by $-1$ in the Young orthogonal form. That vector is cyclic, so the cyclic eigenvector obstruction to invariant vectors excludes invariants.

For sufficiency one can construct an invariant vector explicitly. In a horizontal strip, the occupied column intervals of different rows are disjoint, with every upper interval to the right of every lower interval. Thus an upper-row cell $x$ has content $c(x)$ at least two greater than a lower-row cell $y$ with which it is incomparable. Choose row-reading order as a reference, and for each standard skew Young tableau setOnly pairs in different rows occur in this product; its factors are positive. Put $u={\sum }_T{b}_T{w}_T$. For an admissible swap $R=s_{i}T$ with $d=c_T(i+1)-c_T(i)$, changing the one inversion in the product givesThe two-dimensional Young orthogonal form then fixes $b_T{w}_T+b_R{w}_R$. A nonadmissible swap is within one row and acts by $+1$. Every generator therefore fixes $u\ne 0$, proving existence.

Finally an invariant projection of a cyclic vector generates the invariant subspace, because projection identifies all its translates. That subspace has dimension at most one. This proves the exact multiplicity, including the totally disconnected special case, without asserting the incorrect converse in the PDF.