Show that finitely generated cone in Hilbert space is closed. Let $X$ be a real Hilbert space. Let $\{c_i\}_{i=1}^m\subset X$. Denote
$$
Y=\left\{\sum\limits_{i=1}^m\lambda_ic_i\Big|\lambda_i\in[0,+\infty),\quad 1\leq i\leq m\right\}.
$$
Show that $Y$ is a closed set.
 A: Claim: there is a linearly independent set of vectors  $\{d_i\}_{i=1}^k\subseteq \{c_i\}_{i=1}^m$ that generates $Y$: 
We have  $\tag1y=\sum^m_{i=1}\lambda_ic_i\in Y.$ 
If $\{c_i\}_{i=1}^m$ is already independent, we are done. If not, then 
$\tag2\sum^m_{i=1}z_ic_i=0$ 
where not all the $z_i$ are $0.$ Wlog at least one of the $z_i>0.$  Now, relabel the indices so that $i\ge I\Rightarrow z_i>0$ and $\frac{\lambda_I}{z_I}\ge \cdots \ge \frac{\lambda_m}{z_m}$ for some integer $1\le I\le m$.
Then, combining $(1)$ and $(2),$ we have 
$\tag3 y=\sum_{i=1}^{m-1}\left ( \lambda_i-\frac{z_i}{z_m}\lambda_m\right )c_i.$
Now, $\lambda_i-\frac{z_i}{z_m}\lambda_m\ge 0$ by construction, so we have shown that $y$ can be expressed as a linear combination of $m-1$ vectors. We can repeat this process again, and again, if necessary, and after a finite number of steps we arrive at a linearly independent  $\{d_i\}_{i=1}^k\subseteq \{c_i\}_{i=1}^m.$ 
edit: Following Lin Xuelei's hints in the comments, we can complete the proof: 
We have shown thus far that for each $y\in Y$ there is a linearly independent subset $\left \{ d_i \right \}_i$ of the $\left \{ c_i \right \}_i,\ $ of which $y$ is a linear combination.  
And for each such subset, the map from the subspace generated by the $\left \{ d_i \right \}_i,\ $ i.e. $\sum^k_{i=1}\lambda_id_i\mapsto (\lambda_1,\cdots, \lambda_k)$, is a homeomorphism onto $\mathbb R_{\ge0}^k.$
To finish, notice that $Y$ is a finite disjoint union of all such subspaces, so that in fact, $Y$ is homeomorphic to a finite disjoint union of closed subsets of Euclidean spaces, which is of course closed, which implies now that $Y$ is also closed. 
