Let $V\subseteq\ell^2(\Bbb{Z})$ be the vector space of finitely supported infinite sequences with the induced topology from $\ell^2$ and let $K=\{0\}\cup\{\frac{1}{n}e_n\}_{n\in\Bbb{N}}$, where $e_n$ denotes as usual the vector with $1$ in the $n$th place and $0$ elsewhere. Show that $K$ is compact, and its convex hull is closed but not compact.
Compactness of $K$ was easy to show. Now I tried showing that it is closed but noticed that if I take $$u_n=\frac{\sum_{i=1}^{n} 2^{-i}\frac{e_i}{i}}{\sum_{i=1}^n2^{-i}}$$ $u_n\in \text{co}(K)$ (where $\text{co}(K)$ is the convex hull of $K$) but the limit $$u=\frac{\sum_{i=1}^{\infty}2^{-i}\frac{e_i}{i}}{\sum_{i=1}^\infty2^{-i}}$$ is not (it's not even in the space $V$).
Is there a problem in the question or am I missing something? My suspicion is that I actually proved that $V$ is not closed, but still if $co(K)$ was closed I'd expect that every limit point is in it.
Moreover, I wasnt able to show that it is not compact (not closed would give me that ofcourse, but if it is closed then I don't know how to approach it).
Any hint would be appreciated.
As $u\notin V$, your example has no bearing on whether or not $co(K)$ is closed in $V$. If you had found $u_i\in co(K)$ with $u_i\to u\in V$ but $u\notin co(K)$ then you would have shown that $co(K)$ not closed.
For example $\mathbb{Q}$ is closed in $\mathbb{Q}$, regardless of the fact that a sequence of rationals converges to $\pi$.
However, your example does answer answer another of your questions: it shows that $co(K)$ is not compact. That is, you show that the image of $co(K)$ in $\ell^2(\mathbb{Z})$ (under the inclusion map) is not closed. As $\ell^2(\mathbb{Z})$ is a metric space, we can conclude that the image of $co(K)$ in $\ell^2(\mathbb{Z})$ is not compact. As the image of a compact space is compact, we may conclude that $co(K)$ is not compact.
More explicitly, a cover of $co(K)$ with no finite subcover is given by taking the complement of closed balls of radius $\frac 1n$ about $u$ in $\ell^2(\mathbb{Z})$, and intersecting with $V$.
So in conclusion, whilst your example does not demonstrate the false statement that $co(K)$ is not closed in $V$, your argument is precisely enough to prove the weaker but true statement that $co(K)$ is not compact.
To show $co(K)$ is closed, consider a sequence $u_i$ in $co(K)$, such that $u_i$ converges to $v\in V$. Then for each $j\in \mathbb{N}$ the coefficient $(u_i)_j$ must converge. Let $J$ denote the set of $j\in \mathbb{N}$ such that $(u_i)_j$ converges to a non-zero value. As $v\in V$, we know that $J$ is a finite set.
Let $P:V\to W$ be the natural projection onto the finite dimensional space $W$, spanned by the $e_j|j\in J$. Then $P$ is continuous, so $Pu_i$ converges to $Pv=v$.
As $0\in K$, we know that $x\in co(K)$ if and only if all the coefficients of $x$ are positive, and $\sum_i ix_i\leq1$.
If all the coefficients of $x\in V$ are positive, then all the coefficients of $Px$ will also be positive. If in addition $\sum_i ix_i\leq1$, then $\sum_i i(Px)_i\leq1$.
Thus the $Pu_i\in co(K)\cap W$ and converge to $v\in W$. Thus to show that $v\in co(K)$ it remains to prove that $co(K)\cap W$ is closed in $W$. But this is just the convex hull of a finite set of points in a finite dimensional vector space.