Why are convex polyhedral cones closed?
Here is another proof without induction on dimensions.
Let $A$ be the matrix with column vectors $v_1\dots v_s$.
Claim: The cone $$ \sigma:=\{ x: \ Ay = x, \ y\ge 0\} $$ is closed.
Notation: For an index set $I\subset\{1\dots s\}$ define $\| y \|_I^2 := \sum_{i\in I} y_i^2$.
Proof: Let $(x_k)$ in $\sigma$ be given, such that $x_k\to x$.
The set $\tau_k:=\{ y\ge 0: \ Ay = x_k\}$ is non-empty and closed (the mapping $f:y\mapsto Ay$ is continuous, hence $f^{-1}(\{x\})$ is closed). Hence for each $k$ there is $y_k$ satisfying $$ \|y_k\| = \inf_{y\in \tau_k}\|y\|. $$ If $(y_k)$ does contain a bounded subsequence, then we are done.
It remains to consider the case $\|y_k\|\to \infty$. Denote $\tilde v_k:=\frac1{\|y_k\|}y_k$. By compactness, it contains a converging subsequence. W.l.o.g. let $(\tilde v_k)$ be converging to $v$ with $\|v\|=1$, $v\ge 0$. Moreover, $Av =0$.
Denote $I_1:=\{j: v_j>0\}$, $I_2:=\{1\dots n\}\setminus I_1$. Since $\|v\|=\|v\|_{I_1}$, it follows $\frac{\|y_k\|_{I_1}}{\|y_k\|} \to 1$. Hence, the sequence $(v_k)$ defined by $v_k:=\frac1{\|y_k\|_{I_1}}y_k$ converges to $v$, too. Then there is an index $M$ such that $$ v_{k,i}\in \left[ \frac12 v_i, \frac54 v_i\right] $$ for all $k>M$, $i\in I_1$. In the following let $k>M$. The above inclusion is equivalent to $$ 0\le y_{k,i} - \frac{\|y_k\|_{I_1}}2 v_i\le \frac34 \|y_k\|_{I_1} v_i \quad \forall i\in I_1. $$
Define $z_k:= y_k - \frac{\|y_k\|_{I_1}}2 v$. Then it holds $z_k\ge 0$, $Az_k = x_k$, which implies $z_k\in \tau_k$.
Now, let us show that the norm of $z_k$ is strictly less than the norm of $y_k$, which would give a contradiction to the minimum norm property of $y_k$. We have $$ \begin{split} \|z_k\|^2 & = \|z_k\|_{I_1}^2 + \|z_k\|_{I_2}^2 \le \left(\frac34\|y_k\|_{I_1}\cdot \|v\|_{I_1}\right)^2 + \|y_k\|_{I_2}^2\\ &= \frac9{16} \|y_k\|_{I_1}^2 +\|y_k\|_{I_2}^2 \\ &= \|y_k\|^2 - \frac 7{16}\|y_k\|_{I_1}^2 < \|y_k\|^2, \end{split} $$ which is a contradiction to the minimal norm property of $y_k$.
Note: The claim only holds for a finite set of vectors $v_1\dots v_s$. If $A$ is a linear mapping on an infinite-dimensional space, then the proof fails: We cannot prove that $\frac1{\|y_k\|} y_k$ converges strongly, and we cannot prove that $\frac1{\|y_k\|} y_k$ converges weakly to a non-zero element $v$.
I think this problem has some intrinsic intricacies that you cannot get away from, but a careful induction proof will keep the intricacies down as much as possible.
First, let $W \subset V$ be the unique linear subspace of maximal dimension such that $W \subset \sigma$. Let $q :\mathbb{R}^n \to \mathbb{R}^n / W$ be the quotient by $W$. One checks that $q(\sigma) = \text{Cone}(q(v_1),….,q(v_s))$ and that $\sigma = q^{-1}(q(\sigma))$. This allows one to reduce to the case that $\dim(W)=0$: if this is not true for $\sigma$ in $V$, it is true for $q(\sigma)$ in $V/W$, allowing you to conclude that $q(\sigma)$ is closed, and therefore $\sigma$ is closed by continuity of $q$.
We may therefore assume that $W$ is trivial. It follows that there exists a codimension 1 linear subspace $P$ and a half space $H$ bounded by $P$ such that $\sigma \subset H$ and such that $P \cap \sigma$ is a single point. Let $P'$ be a hyperplane parallel to $P$ that is contained in the interior of $H$. Let $A_i$ be the intersection of $P'$ with the ray $[0,+\infty) \cdot v_i$. Denote the convex hull of the points $A_i$ by $$\text{Hull}(A_1,…,A_s) = \{t_1 A_1 + \cdots + t_s A_s \bigm| t_i \ge 0, t_1 + \cdots + t_s = 1\} \subset P' $$
If we can show that $\text{Hull}(A_1,…,A_s)$ is closed then it is easy to see that $\sigma$ is closed, using that each point of $\sigma$ except for the origin is uniquely represented in the form $r A$ where $r \in (0,+\infty)$ and $A \in \text{Hull}(A_1,….,A_s)$.
In fact the set $\text{Hull}(A_1,…,A_s)$ is compact, so it is closed. Compactness is proved in two steps. First one does one single special case, namely the "standard $s-1$ simplex" which is $$\Delta^{s-1} = \text{Hull}(e_1,…,e_s) = \{(t_1,…,t_s) \in \mathbb{R}^s \bigm| t_i \ge 0, t_1 + \cdots + t_s=1\} $$ where $e_1,…,e_s$ are the standard unit vectors in $\mathbb{R}^s$. Then one uses that to prove compactness of $\text{Hull}(A_1,…,A_s)$ using continuity of the function $\Delta^{s-1} \mapsto \text{Hull}(A_1,…,A_s)$ defined by $(t_1,\ldots,t_s) \mapsto t_1 A_1 + \cdots + t_s A_s$.
Finally some words about compactness of $\Delta^{s-1}$. It is a bounded set, and it is closed because it is the intersection of $s$ closed half-spaces of the hyperplane $P'$.
Added later: Regarding the existence of the half-space $H$ bounded by the hyperplane $P$, here is a proof by induction on dimension. Start by picking an affine half-space that contains $\sigma$ (any proper convex set is contained in a half-space). Translate that half-space to minimize its distance to $\sigma$. We obtain a half-space $H_1$ bounded by the hyperplane $P_1$ such that $\sigma \subset H_1$ and such that $\sigma$ has distance zero from $P_1$. From this one can argue that the origin is contained in $\sigma \cap P_1$. If $\sigma \cap P_1$ is just the origin, we're done. Otherwise, one can argue that $\sigma \cap P_1$ is a face of $\sigma$, meaning the cone of a proper subset of $v_1,…,v_s$, and therefore by induction on dimension $P_1$ contains a halfspace $H_2$ bounded by a hyperplane $P_2$ such that $H_2$ contains the face $\sigma \cap P_1$. Note that $H_2$ has codimension 2 in $V$. We can now rotate $H_1$ around $H_2$ by an arbitrarily small amount so that those of the points $v_1,…,v_s$ that were in the interior $H_1-P_1$ stay in the interior, and those which were in $P_1$ are now in the interior.