A finitely generated cone in Euclidean space is closed

Every $y\in K$ has a representation whose positive-coefficient vectors are linearly independent. Start with any nonnegative coefficient list and discard its zero coefficients. If its remaining vectors are dependent, take a nonzero relation ${\sum }_j{c}_j{a}_j=0$ and change its sign so some $c_j>0$. For the current positive coefficients $t_j$, put $s={\min }_{c_j>0}{t}_j\mathrm{/}{c}_j$. Replacing each $t_j$ by $t_j-s{c}_j$ preserves the represented vector, leaves every coefficient nonnegative, and makes at least one positive coefficient zero. Repeating at most $m$ times leaves an independent positive support.

Fix an independent index set $J$ and let $A_J$ have columns $a_j$, $j\in J$. Its Gram matrix $G_J=A_J^T{A}_J$ is invertible: for every nonzero $v$, $v^T{G}_{J}v=\mathrm{\parallel }{A}_{J}v{\mathrm{\parallel }}_2^2>0$ by independence. Put $B_J=G_J^{-1}{A}_J^T$, define $R_J(y)=B_{J}y$ on ${\mathbb{R}}^n$, and set $L_J=({\sum }_{r=1}^{\mathrm{\mid }J\mathrm{\mid }}{\sum }_{s=1}^n(B_J{)}_{rs}^2{)}^{1\mathrm{/}2}$. Applying [F2] to each row gives $\mathrm{\parallel }{R}_J(y)-R_J(y^{\mathrm{\prime }}){\mathrm{\parallel }}_2^2\le L_J^2\mathrm{\parallel }y-y^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2$, so $R_J$ is Lipschitz with finite constant $L_J$, hence continuous. On the closed subspace $S_J=\mathrm{span}\{a_j:j\in J\}$ it satisfies $A_J{R}_J(y)=y$. Hence $A_J({\mathbb{R}}_{\ge 0}^{\mathrm{\mid }J\mathrm{\mid }})=S_J\cap R_J^{-1}({\mathbb{R}}_{\ge 0}^{\mathrm{\mid }J\mathrm{\mid }})$ is closed: $S_J$ is closed by [F1], the nonnegative orthant is closed, and $R_J$ is continuous. For $J=\mathrm{\varnothing }$ this cone is $\{0\}$, also closed.

By step 1.1, $K$ is the union of the cones in step 1.2 over independent subsets $J\subseteq \{1,\dots ,m\}$. There are only finitely many such subsets, so this union is closed. Each component is a convex cone, and directly taking nonnegative combinations shows $K$ is convex.