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 $v$.
The polyhedral cone $K$ is defined as an intersection of a finite number of half-spaces, i.e. $K=\{x\in\mathbb{R}^n\colon Ax\ge 0\}$, where $A\in\mathbb{R}^{m\times n}$. Since $\text{Im}\,A$ is a subspace, it can be represented as a kernel of some matrix $M$, that is $\ker M=\text{Im} A$. Hence, we have
$$
y=Ax,\ x\in K\qquad\Leftrightarrow\qquad y\in Y=\{y\in\mathbb{R}^m\colon\ My=0,\ y\ge 0\}.\tag1
$$
Introduce the set
$$
P=\{z\in Y\colon\ (\matrix{1 & 1 & \ldots & 1})z=1\}.
$$
It is a bounded polyhedral set, thus, finitely generated (according to what you know)
$$
\exists z_1,z_2,\ldots,z_N\in P\colon\ P=\text{conv}\{z_1,z_2,\ldots,z_N\}.
$$
Therefore, even $Y$ is a finitely generated (positive) cone
$$
Y=\text{cone}\{z_1,z_2,\ldots,z_N\}
$$
since any $y\in Y$ is a non-negative scaling of some $z\in P$.
Now by $(1)$ we can pick $x_k\in K$ such that $z_k=Ax_k$, and we are almost done finding a finite generating set for $K$. The minor trouble left is $\ker A$. Actually, it is quite easy to see that
$$
K=\ker A+\text{cone}\{x_1,x_2,\ldots,x_N\}.
$$
I leave it as an exercise (together with the fact that $\ker A$ is a finitely generated cone).
Best Answer
Indeed, all except perhaps property (2) can be proved without functional analysis. It leads to a completely deterministic algorithm to compute a basis for the dual cone (not a very good algorithm though as its complexity is exponential in the initial basis size).
In the sequel RCP is shorthand for “rational convex polyhedral cone”.
Nearly everything reduces to the iteration of the following lemma:
Fundamental lemma. Let $\sigma={\sf Cone}(u_1,\ldots,u_k)$ be a RPC cone. Let $w\in{{\mathbb Z}^d}$, and $\sigma_w=\lbrace x\in\sigma \,|\, \langle x,w\rangle\ \geq 0\rbrace$. Then $\sigma_w$ is again a RPC cone. In fact, a completely explicit generating set can be found, as follows. Let
$$ \begin{array}{lcl} I_{-} &=& \big\lbrace i \ \big|\ \langle u_i,w\rangle \ < 0 \big\rbrace \\ I_{0} &=& \big\lbrace i \ \big|\ \langle u_i,w\rangle \ = 0 \big\rbrace \\ I_{+} &=& \big\lbrace i \ \big|\ \langle u_i,w\rangle \ < 0 \big\rbrace \\ \end{array} $$
Then $\sigma_w={\sf Cone}({\cal F})$ where
$$ {\cal F}=\big\lbrace u_i\ \big|\ i \in I_0 \cup I_{+}\big\rbrace \cup \big\lbrace \langle u_j,w\rangle u_i-\langle u_i,w\rangle u_j\ \big| \ i \in I_{-}, j\in I_{+}\big\rbrace $$
To show this fundamental lemma we need another :
Lemma 1. Let $n,m \geq 1$ be two integers. Let $x_1,x_2,\ldots,x_n$ and $y_1,y_2,y_3, \ldots ,y_m$ be nonnegative numbers. Then the following are equivalent :
(i) $x_1+x_2+ \ldots +x_n \geq y_1+\ldots +y_m$
(ii) There exists nonnegative numbers $\xi_k(1\leq k\leq n)$ and $t_{ij}(1\leq i \leq n,1\leq j \leq n)$ such that $x_i=\xi_i+\sum_{j}t_{ij}$ for every $i$ and $y_j=\sum_{i}t_{ij}$ for every $j$.
Proof of lemma 1. The direction (ii)$\Rightarrow$(i) is easy because under the hypothesis of (ii) we have
$$ \big(\sum_{i} x_i\big)-\big(\sum_{j} y_j\big)= \big(\sum_{i} \xi_i\big) \geq 0 $$
Conversely, we are going to show (i)$\Rightarrow$(ii) by induction on $N=n+m$. The base case $N=2$ is easy : we have $x_1\geq y_1$ and need just take $\xi_1=0,t_{11}=y_1$.
Now suppose the property is true at level $N$ ; we are going to show it still holds at level $N+1$. So suppose $x_1+x_2+ \ldots +x_n \geq y_1+\ldots +y_m$ for some nonnegative numbers $x_1,x_2,\ldots,x_{n},y_1,y_2,\ldots,y_m$, with $n+m=N+1$. There are two cases.
Case 1. $x_n \geq y_m$. Then we can write $x_n=y_m+X_n$ where $X_n$ is nonnegative. We then have an inequality with one variable less : $x_1+x_2+\ldots+x_{n-1}+X_n \geq y_1+y_2+\ldots +y_{m-1}$ and we can apply the induction hypothesis.
Case 2. $y_m \geq x_n$. Then we can write $y_m=x_n+Y_m$ where $Y_m$ is nonnegative. We then have an inequality with one variable less : $x_1+x_2+\ldots+x_{n-1} \geq y_1+y_2+\ldots +y_{m-1}+Y_m$ and we can apply the induction hypothesis.
This concludes the proof of lemma 1.
Proof of fundamental lemma from lemma 1. Let $\tau={\sf Cone}({\cal F})$. As all vectors in $\cal F$ are nonnegative linear combinations of the $u_i$, whose scalar product with $w$ is nonnegative, we have ${\cal F} \subseteq \sigma_w$ and hence $\tau \subseteq \sigma_w$. Conversely, let $v\in\sigma_w$. Let us put $n=|I_+|,m=|I_{-}|,p=I_{0}$. Reordering the $u_i$ if necessary, we may assume without loss that $I_+=[1,n],I_-=[n+1,n+m],I_0=[n+m,d]$. Then, there are nonnegative coefficients $a_1,a_2,\ldots, a_n,b_1,b_2,\ldots, b_m,c_1,c_2, \ldots, c_p$ such that
$$ v=\sum_{i=1}^{n} a_iu_i- \sum_{j=1}^{m} b_ju_{n+j}+ \sum_{k=1}^{p} c_ku_{n+m+k} \tag{1} $$
Then, the condition $\langle v,w\rangle \geq 0$ can be expressed as
$$ \sum_{i=1}^{n} a_i\langle u_i,w\rangle- \sum_{j=1}^{m} b_j\langle u_{n+j},w\rangle \geq 0 \tag{2} $$
Or, if we put $x_i=a_i\langle u_i,w\rangle$ and $y_j=b_j\langle u_{n+j},w\rangle $,
$$ \sum_{i=1}^{n} x_i- \sum_{j=1}^{m} y_j \geq 0 \tag{3} $$
This is the hypothesis of (i) in lemma 1, so let $\xi_k(1\leq k\leq n)$ and $t_{ij}(1\leq i \leq n,1\leq j \leq n)$ as in (ii) of this lemma. We then have
$$ \begin{array}{lcl} a_i&=&\frac{\xi_i+\sum_{j}t_{ij}}{\langle u_i,w\rangle } (1\leq i\leq n),\\ b_j&=&\frac{\xi_i+\sum_{i}t_{ij}}{\langle u_{n+j},w\rangle } (1\leq j\leq m) \end{array} \tag{4} $$
If we put $z_i=\frac{\xi_i}{\langle u_i,w\rangle }$ and $s_{ij}=\frac{t_{ij}}{\langle u_i,w\rangle\,|\langle u_{n+j},w\rangle|}$, we then have
$$ v=\sum_{i=1}^n z_i u_i +\sum_{i=1}^n \sum_{j=1}^m s_{ij}(\langle u_{n+j},w\rangle u_i-\langle u_i,w\rangle u_{n+j})+ \sum_{k=1}^{p} c_ku_k \tag{5} $$
which is plainly an element of $\tau$. This concludes the proof of the fundamental lemma.
Proof of property (1) from fundamental lemma Notice that ${\cal C}_0={\mathbb R}^d$ is itself a RCP cone (it can be written as ${\sf Cone}(e_1,-e_1,e_2,-e_2, \ldots ,e_d,-e_d)$ where $(e_1,e_2,\ldots ,e_d)$) is the canonical basis of ${\mathbb R}^d$). Then, let
$$ {\cal C}_i=\lbrace v\in {\mathbb R}^d \ | \ \langle v,u_j\rangle \ \geq 0 \ (1\leq j \leq i)\rbrace $$
Iterating the fundamental lemma, we see successively that ${\cal C}_0, {\cal C}_1,\ldots,$ are all RCP cones. In the end ${\sigma}^\vee={\cal C}_k$ is a RCP cone.