Problem : Let $K_1,K_2$ compacts set in $\mathbb{R}^n$. Define : $$ K_1*K_2= \bigcup_{x\in K_1,y \in K_2}[x,y]$$ where $[x,y]=\{ \lambda x + (1-\lambda)y \vert \lambda \in [0,1] \}$
Prove that $K_1*K_2$ is compact.
My proof :
Let $\{a_k\} \in K_1*K_2$, so , for each $k \in \mathbb{N}$ exists $x_k \in K_1$ and $y_k \in K_2$ such that : $a_k \in [x_k,y_k]$. By the definition of $[x_k,y_k]$ exists $\lambda_k \in [0,1]$ such that $a_k = (\lambda_k) x_k + (1-\lambda_k)y_k$. Because $[0,1]$ is compact exists a subsequence $\{\lambda_k\}_{k\in \mathbb{N_1}}$ such that :
$$ \lim_{k\in \mathbb{N}_1} \lambda_k = \lambda \in [0,1] $$
Indeed, because $K_1$ is compact exists a subsequence $\{x_k\}_{k\in\mathbb{N}_2\subseteq \mathbb{N}_1}$ of $\{x_k\}_{k \in \mathbb{N}_1}$ such that :
$$ \lim_{k\in \mathbb{N}_2} x_k = b \in K_1$$
and because $K_2$ is compact exists a subsequence $\{y_k\}_{k\in\mathbb{N}_3 \subseteq \mathbb{N}_2}$ of $\{y_k\}_{k \in \mathbb{N}_2}$ such that :
$$ \lim_{k\in \mathbb{N}_3} y_k = c \in K_2$$
Finally :
$$ \lim_{k \in \mathbb{N}_3}a_k = \lim_{k \in \mathbb{N}_3}(\lambda_k) x_k + \lim_{k \in \mathbb{N}_3}(1-\lambda_k)y_k = \lambda b + (1-\lambda)c=a \in [b,c] $$
So, for each sequence $\{a_k\} \in K_1*K_2$ exists a subsequence convergent to a point of $K_1*K_2$, then $K_1*K_2$ is compact.
I know that my proof is something technical, in several books they resume the part of the use of subsequence saying that :
For the compactness of $K_1$ for example, we can suppose, if it were necessary considering subsequence, that $x_k$ converges to $b$
Everything could be stated in terms of sequence : $\lambda_k, x_k$ and $y_k$ but I wanted to do something more "formal". My proof is correct?
Define $f:[0,1]\times K_1 \times K_2$ by $f(t,x,y)=tx+(1-t)y$. Then $K_1*K_2$ is simply the range of this function. Since $f$ is continuous and its domain is compact, so is the range.