[proofplan] We prove first that one elementary collapse is realized geometrically by a strong deformation retraction. For an elementary collapse removing the interval of faces between a free face $\sigma$ and its unique containing facet $\tau$, the simplex $|\tau|$ deformation retracts onto the union of those faces of $\tau$ that remain in the smaller complex. The free-face condition ensures that this local deformation can be extended by the identity over the rest of the complex. Iterating these strong deformation retractions along a collapse sequence gives a homotopy equivalence from $|K|$ to the realization of a one-vertex complex, which is a point. [/proofplan]

[step:Reduce the theorem to invariance under one elementary collapse]Let $K = K_0 \searrow K_1 \searrow \cdots \searrow K_m$ be a collapse sequence such that $K_m = \{\varnothing, \{v\}\}$ for some vertex $v$. Here each $K_{j+1}$ is obtained from $K_j$ by an elementary collapse in the free-face sense: there are faces $\sigma_j \subsetneq \tau_j$ of $K_j$ such that $\tau_j$ is a facet of $K_j$, $\sigma_j$ is contained in no facet of $K_j$ other than $\tau_j$, and $K_{j+1}$ is obtained by deleting all faces $\rho$ with $\sigma_j \subseteq \rho \subseteq \tau_j$. It is enough to prove that, for every $j \in \{0,\dots,m-1\}$, the inclusion map $i_j: |K_{j+1}| \to |K_j|$ is a homotopy equivalence. Indeed, the composite of these homotopy equivalences identifies $|K| = |K_0|$ up to homotopy with $|K_m|$, and $|K_m|$ consists of the single point corresponding to $v$.[/step]

[guided]The collapsibility hypothesis gives more than an abstract homotopy equivalence: it gives a finite list of explicit elementary collapses. Write this list as $K = K_0 \searrow K_1 \searrow \cdots \searrow K_m$, where the last complex is the one-vertex complex $K_m = \{\varnothing, \{v\}\}$. Thus the proof only needs one local statement: whenever $K_{j+1}$ is obtained from $K_j$ by one elementary collapse, the space $|K_j|$ deformation retracts onto $|K_{j+1}|$. If this is true, then each inclusion $i_j: |K_{j+1}| \to |K_j|$ is a homotopy equivalence. Composing the resulting homotopy equivalences gives that $|K| = |K_0|$ is homotopy equivalent to $|K_m|$. Since $|K_m|$ is a one-point space, $|K|$ is contractible.[/guided]

[step:Describe the elementary collapse inside one simplex] Let $L$ be a finite simplicial complex, and let $L'$ be obtained from $L$ by an elementary collapse. Thus there are faces $\sigma \subsetneq \tau$ of $L$ such that $\tau$ is a facet of $L$, $\sigma$ is contained in no facet of $L$ other than $\tau$, and $L' = L \setminus \{\rho \in L : \sigma \subseteq \rho \subseteq \tau\}$. Let $V(\tau)$ denote the finite vertex set of $\tau$, and let $V(\sigma) \subset V(\tau)$ denote the vertex set of $\sigma$. Define the remaining part of the simplex $|\tau|$ by $A := |\tau| \cap |L'|$. In barycentric coordinates on $|\tau|$, this set is $A = \{x \in |\tau| : \text{there exists } a \in V(\sigma) \text{ with } x_a = 0\}$. Indeed, a point $x \in |\tau|$ lies in the relative interior of the unique face $\rho_x := \{w \in V(\tau) : x_w > 0\}$. The point $x$ belongs to the deleted part exactly when $\sigma \subseteq \rho_x \subseteq \tau$, equivalently when $x_a > 0$ for every $a \in V(\sigma)$. [/step]

[step:Construct a deformation retraction of the collapsed simplex]Choose a vertex $b \in V(\tau) \setminus V(\sigma)$, which exists because $\sigma \subsetneq \tau$. Define the map $p: |\tau| \to A$ as follows. Define the function $\lambda: |\tau| \to [0,1]$ by $\lambda(x) := \min_{a \in V(\sigma)} x_a$. The barycentric coordinates of $p(x)$ are defined by \begin{align*} p(x)_a := x_a - \lambda(x) \end{align*} for $a \in V(\sigma)$, \begin{align*} p(x)_b := x_b + |V(\sigma)|\lambda(x), \end{align*} and \begin{align*} p(x)_w := x_w \end{align*} for $w \in V(\tau) \setminus (V(\sigma) \cup \{b\})$. These coordinates are non-negative and have total sum $1$, so $p(x) \in |\tau|$. Since at least one coordinate $p(x)_a$ with $a \in V(\sigma)$ is zero, we have $p(x) \in A$. If $x \in A$, then $\lambda(x)=0$, hence $p(x)=x$. Define $H_\tau: |\tau| \times [0,1] \to |\tau|$ by $H_\tau(x,t) := (1-t)x + t p(x)$. For every $x \in |\tau|$ and $t \in [0,1]$, the barycentric coordinates of $H_\tau(x,t)$ are non-negative and sum to $1$, so $H_\tau(x,t) \in |\tau|$. Moreover $H_\tau(x,0)=x$, $H_\tau(x,1)=p(x) \in A$, and $H_\tau(x,t)=x$ for all $x \in A$. Thus $H_\tau$ is a strong deformation retraction of $|\tau|$ onto $A$.[/step]

[guided]We now build the local deformation explicitly. Since $\sigma$ is a proper face of $\tau$, there is at least one vertex of $\tau$ not belonging to $\sigma$; choose one and call it $b \in V(\tau) \setminus V(\sigma)$. The set $A$ is the part of $|\tau|$ that remains after deleting the faces containing $\sigma$. In barycentric coordinates, a point belongs to $A$ exactly when at least one coordinate indexed by a vertex of $\sigma$ is zero. Therefore, to push a point of $|\tau|$ into $A$, it is enough to reduce all $\sigma$-coordinates by their common minimum. Define the function $\lambda: |\tau| \to [0,1]$ by $\lambda(x) := \min_{a \in V(\sigma)} x_a$ for $x \in |\tau|$. Now define a map $p: |\tau| \to A$ by modifying barycentric coordinates as follows: \begin{align*} p(x)_a := x_a - \lambda(x) \end{align*} for every $a \in V(\sigma)$, \begin{align*} p(x)_b := x_b + |V(\sigma)|\lambda(x), \end{align*} and \begin{align*} p(x)_w := x_w \end{align*} for every remaining vertex $w \in V(\tau) \setminus (V(\sigma) \cup \{b\})$. The coordinates are non-negative because $\lambda(x) \le x_a$ for every $a \in V(\sigma)$. Their sum is still $1$: we subtract $\lambda(x)$ once for each of the $|V(\sigma)|$ vertices of $\sigma$, and then add exactly $|V(\sigma)|\lambda(x)$ to the coordinate of $b$. Hence $p(x)$ is a point of $|\tau|$. At least one $a \in V(\sigma)$ satisfies $x_a = \lambda(x)$, and for this vertex $p(x)_a = 0$; therefore $p(x) \in A$. If $x \in A$, then some coordinate $x_a$ with $a \in V(\sigma)$ is already zero. Since all barycentric coordinates are non-negative, this gives $\lambda(x)=0$, and therefore $p(x)=x$. Thus $p$ fixes $A$ pointwise. Finally define $H_\tau: |\tau| \times [0,1] \to |\tau|$ by $H_\tau(x,t) := (1-t)x + t p(x)$. This is the straight-line homotopy in the affine simplex $|\tau|$. Convexity of $|\tau|$ ensures that $H_\tau(x,t) \in |\tau|$ for all $x \in |\tau|$ and $t \in [0,1]$. We have $H_\tau(x,0)=x$, $H_\tau(x,1)=p(x) \in A$, and if $x \in A$, then $p(x)=x$, so $H_\tau(x,t)=x$ for every $t$. Hence $H_\tau$ is a strong deformation retraction of $|\tau|$ onto $A$.[/guided]

[step:Extend the local deformation over the whole complex]Define $H: |L| \times [0,1] \to |L|$ by \begin{align*} H(x,t) := H_\tau(x,t) \end{align*} when $x \in |\tau|$, and by \begin{align*} H(x,t) := x \end{align*} when $x \in |L'|$. This is well-defined because the two formulas agree on $|\tau| \cap |L'| = A$, where $H_\tau$ fixes every point. Since $|L| = |L'| \cup |\tau|$ and both $|L'|$ and $|\tau|$ are closed subspaces of the finite polyhedron $|L|$, continuity of $H$ follows from the closed-cover gluing criterion for continuous maps: the restrictions of $H$ to $|L'| \times [0,1]$ and $|\tau| \times [0,1]$ are continuous and agree on their intersection. The map satisfies $H(x,0)=x$ for all $x \in |L|$, satisfies $H(x,t)=x$ for all $x \in |L'|$, and satisfies $H(x,1) \in |L'|$ for all $x \in |L|$. Therefore $H$ is a strong deformation retraction of $|L|$ onto $|L'|$.[/step]

[guided]The local homotopy $H_\tau$ has to be patched with the identity outside the simplex being collapsed. Define $H: |L| \times [0,1] \to |L|$ by using $H_\tau$ on $|\tau|$ and the identity on $|L'|$. The only possible ambiguity occurs at points of $|\tau| \cap |L'|$, and this intersection is exactly $A$. Since $H_\tau$ fixes every point of $A$, the two formulas agree there, so $H$ is well-defined. We next check continuity. The realization $|L|$ is the union $|L'| \cup |\tau|$. Because $L$ is finite, $|L'|$ and $|\tau|$ are closed subspaces of the finite polyhedron $|L|$. Hence $|L'| \times [0,1]$ and $|\tau| \times [0,1]$ are closed subspaces covering $|L| \times [0,1]$. On these two closed pieces, the restrictions of $H$ are respectively the identity homotopy and the continuous map $H_\tau$; they agree on the overlap because $H_\tau$ fixes $A$. The closed-cover gluing criterion therefore gives that $H$ is continuous. Finally, $H(x,0)=x$ for every $x \in |L|$. If $x \in |L'|$, then $H(x,t)=x$ for every $t \in [0,1]$. If $x \in |\tau|$, then $H(x,1)=H_\tau(x,1) \in A \subset |L'|$, while if $x \in |L'|$ then $H(x,1)=x \in |L'|$. Thus $H(-,1)$ lands in $|L'|$ and fixes $|L'|$ pointwise throughout the homotopy. This proves that $H$ is a strong deformation retraction of $|L|$ onto $|L'|$.[/guided]

[step:Iterate the deformation retractions to reach a point]Applying the previous step to each elementary collapse $K_j \searrow K_{j+1}$, the space $|K_j|$ strongly deformation retracts onto $|K_{j+1}|$ for every $j \in \{0,\dots,m-1\}$. Hence $|K|=|K_0|$ is homotopy equivalent to $|K_m|$. Since $K_m$ is the one-vertex complex, $|K_m|$ is a singleton. Thus $|K|$ is homotopy equivalent to a point, so $|K|$ is contractible.[/step]

[guided]We have proved the required local statement: every elementary collapse $K_j \searrow K_{j+1}$ gives a strong deformation retraction from $|K_j|$ onto $|K_{j+1}|$. In particular, the inclusion $|K_{j+1}| \to |K_j|$ is a homotopy equivalence for each $j \in \{0,\dots,m-1\}$. Composing these homotopy equivalences along the finite collapse sequence gives a homotopy equivalence from $|K|=|K_0|$ to $|K_m|$. The terminal complex is $K_m=\{\varnothing,\{v\}\}$, so its realization consists of the single point corresponding to the vertex $v$. A space homotopy equivalent to a point is contractible. Therefore $|K|$ is contractible.[/guided]