Let $X\subseteq\Bbb R^2$, $Y$ a metric space, and $f:X\to Y$. Suppose $f$ is uniformly continuous on every bounded set $B\subseteq X$. Then there is an extension $g:\overline{X}\to Y$ such that $g$ is continous and $g\vert_X=f$. How to prove that?
Can I say, for every point $a\in X$, choose $B_r^X(a)=\{x\in X\mid d(x,a)<r\}$ be a bounded set, then since $f$ is uniformly continuous on it, we can continuously extend it to $B_r^X(a)\cup\partial B_r^X(a)$?
Assuming that Y is complete here is the proof: let $x \in \bar {X}$ and choose a sequence $\{x_n\}$ in X converging to x. Then $\{x_n:n=1,2,...\}$ is a bounded set so f is uniformly continuous on this set. It follows easily from tthe definition of uniform continuity and the fact that $||x_n -x_m|| \to 0$ as $n,m \to \infty$ that $D(f(x_n),f(x_m)) \to 0$ where $D$ is the metric on Y. By completeness there exists $y \in Y$ such that $D(f(x_n),y) \to 0$. Define $f(x)$ as $y$. If you take another sequence $\{y_n\}$ in X converging to x then $||x_n -y_n|| \to 0$ and this gives $D(f(x_n) ,f(y_n)) \to 0$ because f is uniformly continuous on $\{x_1,y_1,x_2,y_2,\cdots\}$ Hence f is well defined. If $x \in X$ then we can take $x_n =x$ for all $n$ which shows that the function we have defined is an extension of f. The extended function is also continuous but I am not proving it because you did not state continuity!