If $K$ is compact and $C$ is closed in $\mathbb{R}^k$, prove that $K + C$ is closed using a "direct" proof Rudin Exercise 4.25(a) reads: 
If $K$ is compact and $C$ is closed in $\mathbb{R}^k$, prove that $K + C$ is closed.
The hints in the problem suggest a proof by proving that the complement of $K + C$ is open, a path which I was able to follow into a successful proof.  However, I want to prove it "directly", by showing that any limit point of $K + C$ must be within $K + C$, however I run into the following problem:
My Attempt:
Suppose $z$ is a limit point of $K + C$.  Then there is a sequence $\{z_n\} \to z$ in $K + C$.  Since each $z_i$ is an element of $K + C$, we can write $z_i = k_i + c_i$ for sequences $\{k_i\}$, $\{c_i\}$ in $K$ and $C$ respectively. Now, we simply must show that $\{k_i\} \to k \in K$ and $\{c_i\} \to c \in C$ to be done. 
However, I noticed that $\{k_n\}$ and $\{c_n\}$ do not necessarily converge when their sum does. As an example, in $\mathbb{R}$, take $k_i = (-1)^i$ and $c_i = (-1)^{i+1}$.  Then neither $\{k_n\}$ nor $\{c_n\}$ converge, but their sum does.
Are there any suggestions on how to get around this problem and complete this more "direct" proof?
Thanks!
 A: As $K$ is compact, there is a subsequence $k_{i_j}$ such that coverges to  some $k \in K$. In particular, the subsequence $z_{i_j} \rightarrow z$. Thus, $c_{i_j}= z_{i_j} -k_{i_j} \rightarrow z - k \in C$, since $C $ is closed. Then, 
$$ 
z= k + (z-k) \in K + C.
$$
A: There does exist a proof using your start. Your missing step was fully using the compactness of $K$:
Let $z$ be any limit point of $K + C$. Then there exists an infinite sequence of points $\{z_n\} \in K + C$ that converges to $z$. Define sequences $\{k_n\} \in K$ and $\{c_n\} \in C$ such that $z_n = k_n + c_n$ for all $n$. $\{k_n\}$ is an infinite sequence in a compact space, therefore $\{k_n\}$ must have some limit point $k \in K$. Thus for any $\varepsilon > 0$, there is an index $n$ for which $|k - k_n| < \varepsilon / 2$ and $|k_n + c_n - z| < \varepsilon / 2$. Put $c = z - k$. By the triangle inequality, we have $|c_n - c| < \varepsilon$ at this $n$. Hence $c$ is a limit point of $\{c_n\}$. This shows that $z = k+c$ for $k \in K$ and $c \in C$; therefore $K + C$ is closed.
