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Once more, with feeling. Thanks to comments from Tyler Lawson and Neil Strickland.

Let me use $B_n$ for the finite Boolean lattice of subsets of $[n] := \{1,2,\ldots,n\}$ (your $\mathcal{P}(S)$), and let $k=|T|$. Then your $A=((B_k\setminus \{\hat{1}\}) \times B_{n-k}) \setminus \{\hat{0}\}$, where $\hat{0}$ means the minimum of a poset, and $\hat{1}$ the maximum. To understand the homotopy type of this product, we need a result of Quillen (stated as Theorem 5.1(b) in the paper "Canonical homeomorphisms of posets" by Walker cited below; see also the discussion in Section 5.1 of the notes of Wachs), which says that $$ |P\times Q \setminus \{\hat{0}\}| \simeq |P\setminus \{\hat{0}\}| \ast |Q\setminus\{\hat{0}\}|,$$ for posets $P,Q$ which have minimums, where $\ast$ is the join of complexes. So in our case $$ |A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \ast |B_{n-k}\setminus \{\hat{0}\}|.$$ But $B_{n-k}\setminus \{\hat{0}\}$ is contractible (it has a maximum), so $|A|$ is also contractible.

Note that if we considered the poset $B=\{U\subseteq [n]\colon U \cap T \neq \varnothing, T \not\subseteq U\}$ as in the answer of Neil Strickland, then we have that $B = (B_k \setminus \{\hat{0},\hat{1}\}) \times B_{n-k}$. [And this was the poset that my first answer posted here was implicitly about.] Using the another fact about homotopy types of products (see Theorem 5.1(a) in the paper of Walker) that $$|P\times Q| \simeq |P| \times |Q|$$ (product of complexes) means that $$|B| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \times |B_{n-k}|,$$ and since $B_{n-k}$ is contractible, this time we actually can say $|B| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \simeq \mathbb{S}^{k-2}$.

Quillen, Daniel, Homotopy properties of the poset of nontrivial p-subgroups of a group, Adv. Math. 28, 101-128 (1978). ZBL0388.55007.

Wachs, Michelle L., Poset topology: tools and applications, Miller, Ezra (ed.) et al., Geometric combinatorics. Providence, RI: American Mathematical Society (AMS); Princeton, NJ: Institute for Advanced Studies (ISBN 978-0-8218-3736-8/hbk). IAS/Park City Mathematics Series 13, 497-615 (2007). ZBL1135.06001.

Walker, James W., Canonical homeomorphisms of posets, Eur. J. Comb. 9, No. 2, 97-107 (1988). ZBL0661.06006.

Once more, with feeling. Thanks to comments from Tyler Lawson and Neil Strickland.

Let me use $B_n$ for the finite Boolean lattice of subsets of $[n] := \{1,2,\ldots,n\}$ (your $\mathcal{P}(S)$), and let $k=|T|$. Then your $A=\{U\subseteq[n]\colon U\neq \varnothing, T \not \subseteq U\}=((B_k\setminus \{\hat{1}\}) \times B_{n-k}) \setminus \{\hat{0}\}$, where $\hat{0}$ means the minimum of a poset, and $\hat{1}$ the maximum. To understand the homotopy type of this product, we need a result of Quillen (stated as Theorem 5.1(b) in the paper "Canonical homeomorphisms of posets" by Walker cited below; see also the discussion in Section 5.1 of the notes of Wachs), which says that $$ |P\times Q \setminus \{\hat{0}\}| \simeq |P\setminus \{\hat{0}\}| \ast |Q\setminus\{\hat{0}\}|,$$ for posets $P,Q$ which have minimums, where $\ast$ is the join of topological spaces. So in our case $$ |A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \ast |B_{n-k}\setminus \{\hat{0}\}|.$$ But $B_{n-k}\setminus \{\hat{0}\}$ is contractible (it has a maximum), so $|A|$ is also contractible.

Note that if we considered the poset $B=\{U\subseteq [n]\colon U \cap T \neq \varnothing, T \not\subseteq U\}$ as in the answer of Neil Strickland, then we have that $B = (B_k \setminus \{\hat{0},\hat{1}\}) \times B_{n-k}$. [This is the poset that the first version of my answer was implicitly about.] Here we need a slightly different fact about homotopy types of products, namely that $$|P\times Q| \simeq |P| \times |Q|,$$ where $\times$ means product of topological spaces (see Theorem 5.1(a) in the paper of Walker, or again Section 5.1 of the notes of Wachs). This means that $$|B| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \times |B_{n-k}|,$$ and since $B_{n-k}$ is contractible, this time we actually can say $|B| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \simeq \mathbb{S}^{k-2}$.

Quillen, Daniel, Homotopy properties of the poset of nontrivial p-subgroups of a group, Adv. Math. 28, 101-128 (1978). ZBL0388.55007.

Wachs, Michelle L., Poset topology: tools and applications, Miller, Ezra (ed.) et al., Geometric combinatorics. Providence, RI: American Mathematical Society (AMS); Princeton, NJ: Institute for Advanced Studies (ISBN 978-0-8218-3736-8/hbk). IAS/Park City Mathematics Series 13, 497-615 (2007). ZBL1135.06001.

Walker, James W., Canonical homeomorphisms of posets, Eur. J. Comb. 9, No. 2, 97-107 (1988). ZBL0661.06006.

I've corrected this answer now, thanks to the comment from Tyler Lawson.

Let me use $B_n$ for the finite Boolean lattice of subsets of $[n] := \{1,2,\ldots,n\}$ (your $\mathcal{P}(S)$), and let $k=|T|$. Then your $A=((B_k\setminus \{\hat{1}\}) \times B_{n-k}) \setminus \{\hat{0}\}$, where $\hat{0}$ means the minimum of a poset, and $\hat{1}$ the maximum. To understand the homotopy type of this product, we need a result of Quillen (stated as Theorem 5.1(b) in the paper "Canonical homeomorphisms of posets" by Walker cited below; see also the discussion in Section 5.1 of the notes of Wachs), which says that $$ |P\times Q \setminus \{\hat{0}\}| \simeq |P\setminus \{\hat{0}\}| \ast |Q\setminus\{\hat{0}\}|,$$ for posets $P,Q$ which have minimums, where $\ast$ is the join of complexes. So in our case $$ |A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \ast |B_{n-k}\setminus \{\hat{0}\}|.$$ But $B_{n-k}\setminus \{\hat{0}\}$ is contractible (it has a maximum), so $|A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \simeq \mathbb{S}^{k-2}$, as you guessed.

Quillen, Daniel, Homotopy properties of the poset of nontrivial p-subgroups of a group, Adv. Math. 28, 101-128 (1978). ZBL0388.55007.

Wachs, Michelle L., Poset topology: tools and applications, Miller, Ezra (ed.) et al., Geometric combinatorics. Providence, RI: American Mathematical Society (AMS); Princeton, NJ: Institute for Advanced Studies (ISBN 978-0-8218-3736-8/hbk). IAS/Park City Mathematics Series 13, 497-615 (2007). ZBL1135.06001.

Walker, James W., Canonical homeomorphisms of posets, Eur. J. Comb. 9, No. 2, 97-107 (1988). ZBL0661.06006.

Once more, with feeling. Thanks to comments from Tyler Lawson and Neil Strickland.

Let me use $B_n$ for the finite Boolean lattice of subsets of $[n] := \{1,2,\ldots,n\}$ (your $\mathcal{P}(S)$), and let $k=|T|$. Then your $A=((B_k\setminus \{\hat{1}\}) \times B_{n-k}) \setminus \{\hat{0}\}$, where $\hat{0}$ means the minimum of a poset, and $\hat{1}$ the maximum. To understand the homotopy type of this product, we need a result of Quillen (stated as Theorem 5.1(b) in the paper "Canonical homeomorphisms of posets" by Walker cited below; see also the discussion in Section 5.1 of the notes of Wachs), which says that $$ |P\times Q \setminus \{\hat{0}\}| \simeq |P\setminus \{\hat{0}\}| \ast |Q\setminus\{\hat{0}\}|,$$ for posets $P,Q$ which have minimums, where $\ast$ is the join of complexes. So in our case $$ |A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \ast |B_{n-k}\setminus \{\hat{0}\}|.$$ But $B_{n-k}\setminus \{\hat{0}\}$ is contractible (it has a maximum), so $|A|$ is also contractible.

Note that if we considered the poset $B=\{U\subseteq [n]\colon U \cap T \neq \varnothing, T \not\subseteq U\}$ as in the answer of Neil Strickland, then we have that $B = (B_k \setminus \{\hat{0},\hat{1}\}) \times B_{n-k}$. [And this was the poset that my first answer posted here was implicitly about.] Using the another fact about homotopy types of products (see Theorem 5.1(a) in the paper of Walker) that $$|P\times Q| \simeq |P| \times |Q|$$ (product of complexes) means that $$|B| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \times |B_{n-k}|,$$ and since $B_{n-k}$ is contractible, this time we actually can say $|B| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \simeq \mathbb{S}^{k-2}$.

Quillen, Daniel, Homotopy properties of the poset of nontrivial p-subgroups of a group, Adv. Math. 28, 101-128 (1978). ZBL0388.55007.

Wachs, Michelle L., Poset topology: tools and applications, Miller, Ezra (ed.) et al., Geometric combinatorics. Providence, RI: American Mathematical Society (AMS); Princeton, NJ: Institute for Advanced Studies (ISBN 978-0-8218-3736-8/hbk). IAS/Park City Mathematics Series 13, 497-615 (2007). ZBL1135.06001.

Walker, James W., Canonical homeomorphisms of posets, Eur. J. Comb. 9, No. 2, 97-107 (1988). ZBL0661.06006.

I've corrected this answer now, thanks to the comment from Tyler Lawson.

Let me use $B_n$ for the finite Boolean lattice of subsets of $[n] := \{1,2,\ldots,n\}$ (your $\mathcal{P}(S)$), and let $k=|T|$. Then your $A=((B_k\setminus \{\hat{1}\}) \times B_{n-k}) \setminus \{\hat{0}\}$, where $\hat{0}$ means the minimum of a poset, and $\hat{1}$ the maximum. To understand the homotopy type of this product, we need a result of Walker (Theorem 5.1(b) in "Canonical homeomorphisms of posets" cited below; see also the discussion in Section 5.1 of the notes of Wachs), which says that $$ |P\times Q \setminus \{\hat{0}\}| \simeq |P\setminus \{\hat{0}\}| \ast |Q\setminus\{\hat{0}\}|,$$ for posets $P,Q$ which have minimums, where $\ast$ is the join of complexes. So in our case $$ |A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \ast |B_{n-k}\setminus \{\hat{0}\}|.$$ But $(B_{n-k}\setminus \{\hat{0}\})$ is contractible (it has a maximum), so $|A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \simeq \mathbb{S}^{k-2}$, as you guessed.

Wachs, Michelle L., Poset topology: tools and applications, Miller, Ezra (ed.) et al., Geometric combinatorics. Providence, RI: American Mathematical Society (AMS); Princeton, NJ: Institute for Advanced Studies (ISBN 978-0-8218-3736-8/hbk). IAS/Park City Mathematics Series 13, 497-615 (2007). ZBL1135.06001.

Walker, James W., Canonical homeomorphisms of posets, Eur. J. Comb. 9, No. 2, 97-107 (1988). ZBL0661.06006.

I've corrected this answer now, thanks to the comment from Tyler Lawson.

Let me use $B_n$ for the finite Boolean lattice of subsets of $[n] := \{1,2,\ldots,n\}$ (your $\mathcal{P}(S)$), and let $k=|T|$. Then your $A=((B_k\setminus \{\hat{1}\}) \times B_{n-k}) \setminus \{\hat{0}\}$, where $\hat{0}$ means the minimum of a poset, and $\hat{1}$ the maximum. To understand the homotopy type of this product, we need a result of Quillen (stated as Theorem 5.1(b) in the paper "Canonical homeomorphisms of posets" by Walker cited below; see also the discussion in Section 5.1 of the notes of Wachs), which says that $$ |P\times Q \setminus \{\hat{0}\}| \simeq |P\setminus \{\hat{0}\}| \ast |Q\setminus\{\hat{0}\}|,$$ for posets $P,Q$ which have minimums, where $\ast$ is the join of complexes. So in our case $$ |A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \ast |B_{n-k}\setminus \{\hat{0}\}|.$$ But $B_{n-k}\setminus \{\hat{0}\}$ is contractible (it has a maximum), so $|A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \simeq \mathbb{S}^{k-2}$, as you guessed.

Quillen, Daniel, Homotopy properties of the poset of nontrivial p-subgroups of a group, Adv. Math. 28, 101-128 (1978). ZBL0388.55007.

Wachs, Michelle L., Poset topology: tools and applications, Miller, Ezra (ed.) et al., Geometric combinatorics. Providence, RI: American Mathematical Society (AMS); Princeton, NJ: Institute for Advanced Studies (ISBN 978-0-8218-3736-8/hbk). IAS/Park City Mathematics Series 13, 497-615 (2007). ZBL1135.06001.

Walker, James W., Canonical homeomorphisms of posets, Eur. J. Comb. 9, No. 2, 97-107 (1988). ZBL0661.06006.