Compact set plus a closed set is a closed set Let $(X,\|\;\|)$ be a normed vector space over $K$ and let $A,B\subseteq X$ be s.t. $A$ is compact and $B$ is closed. I want to prove that:
$$(\alpha A+\beta B)\;\text{ is closed }\;\forall \alpha,\beta\in K$$
So, given our function $\odot$ as in here, my attempt goes like this:
Let $\alpha,\beta\in K$. If $\alpha=\beta=0$, clearly
$$\alpha A+\beta B=\{0\}\text{ is closed}$$
Now, since $\odot$ is continuous and $A$ is compact, we have that $\;\forall \alpha\neq0,\;\alpha A=\odot(\alpha,A)\;$ is compact and thus closed.

Then, let $\;y\in \beta B$ so, 
$$
y=\beta b_y
$$
for some $b_y\in B$. But, since $B$ is closed, there exists $\{b_n\}_{n\in\Bbb N}\subset B$ s.t. $b_n\xrightarrow[n\to\infty]{}b_y$
So let $y_n=\beta b_n\;\;\forall n\in\Bbb N$. Thus, $\{y_n\}_{n\in\Bbb N}\subset \beta B$ and $\|y_n-y\|=\|\beta b_n-\beta b_y\|=|\beta|\|b_n-b_y\|\xrightarrow[n\to\infty]{}0$
Thus, we get that 
$$\forall y\in\beta B\;\exists\;\{y_n\}_{n\in\Bbb N}\subset \beta B \text{  s.t.  } y_n\xrightarrow[n\to\infty]{}y$$
so $\beta B$ is closed $\;\forall \beta\in K$.
Now,
$$
\text{Let }x\in (\alpha A+\beta B)\Rightarrow\;\exists\;a^*\in\alpha A\;\land\;b^*\in\beta B\text{  s.t.  } x=a^*+b^*
$$
so, since $\alpha A$ and $\beta B$ are closed,
$$
\exists\;\{a_n^*\}_{n\in\Bbb N}\subset\alpha A\text{  s.t.  }a_n^*\xrightarrow[n\to\infty]{}a^*\\
\exists\;\{b_n^*\}_{n\in\Bbb N}\subset\beta B\text{  s.t.  }b_n^*\xrightarrow[n\to\infty]{}b^*
$$
so, taking $x_n=a_n^*+b_n^*\;\forall n\in\Bbb N$ is clear that $x_n\in (\alpha A+\beta B)\;\forall n\in\Bbb N$ and
$$
\|x_n-x\|=\|a_n^*+b_n^*-a^*-b^*\|\le\|a_n^*-a^*\|+\|b_n^*-b^*\|\xrightarrow[n\to\infty]{}0
$$
so $x_n\xrightarrow[n\to\infty]{}x$ and, since this happens $\;\forall x\in (\alpha A+\beta B)$, thereby $(\alpha A+\beta B)$ is closed $\;\forall \alpha ,\beta\in K$
What do you think? Did I miss something?
 A: You do not start correctly when you want to prove that a set is closed. When you want to prove that $A\subset X$ is closed, don't choose $x\in A$ and show that there is a sequence $(x_n)$ in $A$ that converges to $x$, since this sequence might be $x_n = x$. In contrast, you should start with a convergent sequence $(x_n)$ in $A$ and show that its limit also is an element of $A$.
In your case, $\beta B$ is closed. Clearly, this holds for $\beta = 0$. So, let $\beta\neq 0$. Choose $(x_n)\subset\beta B$, $x_n\to x$. Then $x_n = \beta b_n$ with $b_n\in B$. Now, $b_n = \beta^{-1}x_n\to\beta^{-1}x$. As $B$ is closed, $\beta^{-1}x\in B$, so $x\in\beta B$. Done.
Here, as an alternative, you can also define $f : X\to X$ by $f(x) := \beta^{-1}x$. Then $\beta B = f^{-1}(B)$. And that was it since $f$ is continuous.
Your proof that $\alpha A$ is compact is fine. Now, apply the above approach to show that $C + D$ is closed if $C$ is compact and $D$ closed. Use here that each sequence in $C$ has a convergent subsequence.
