Let's fix a triangle
$\Delta=\{t_1x_1+t_2x_2+t_3x_3 \in \mathbb{R}^2 \mid t_1,t_2,t_3 \ge 0 \quad \land \quad t_1+t_2+t_3=1\}$
of fixed vertices $x_1,x_2,x_3 \in \mathbb{R}^2$.
I want to show that $\Delta$ is compact in the plane. That's my attempt (the metric used here is the euclidean one, of course):
Let's define $f\colon \mathbb{R}^3 \to \mathbb{R}^2 \mid f(t_1,t_2,t_3)=t_1x_1+t_2x_2+t_3x_3$,
$g\colon \mathbb{R}^3 \to \mathbb{R} \mid g(t_1,t_2,t_3)=t_1+t_2+t_3$.
Obviously $f$ and $g$ are continuous functions.
So the set:
$K=g^{-1}(\{1\}) \cap \{(t_1,t_2,t_3) \in \mathbb{R}^3 \mid t_1,t_2,t_3 \ge 0\}$
is closed, being the intersection of two closed sets.
Moreover:
$\forall \,(t_1,t_2,t_3) \in K \quad \|(t_1,t_2,t_3)-(0,0,0)\|=\sqrt{t_1^2+t_2^2+t_3^2} \le t_1+t_2+t_3=1<2$
so that $K$ is also bounded.
So $K$ is compact, and thus $\Delta=f(K)$ is compact by Weierstrass theorem.
Is it correct? Is there a easier (elementary) way to prove it? Thank you!
Your method works, though I am not sure which theorem you have in mind when you say "by Weierstrass theorem." The image of a compact set under a continuous function is compact, so the triangle is compact, like you showed.
Can you prove that the line segment in $\mathbb R^2$ joining two points $P$ and $Q$ is closed? If so, then you have a direct proof that triangles (in fact any polygons) are closed, being a union of a finite number of such segments.