If $K$ is compact and $C$ is closed in $\mathbb{R}^k$, prove that $K + C$ is closed using a "direct" proof

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, $$ z= k + (z-k) \in K + C. $$

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.