Prove there exists a unique continuous function suppose that $f:A\to\mathbb R$ is uniformly continuous on $A$. Let $\overline{A}$ be the closure of $A$.
Assuming that there exists a continuous extension of $f$,  
$g:\overline{A}\to \mathbb R$ s.t. for all $x \in A, g(x)=f(x)$ 
prove that the extension is unique.
I'm fairly certain that I have to assume that there are two functions, say $g$ and $g'$, and somehow conclude that they are actually the same function. I'm quite confused as how to go further. Any suggestions or help is appreciated. Thank you in advance! 
 A: Suppose that $h$ is a continuous extension of $f$ to $\bar{A}$ so that $f(x)=h(x)$ on $A$.  We want to show that $h(x)=g(x)$ for all $x\in \bar{A}$.
Let $y$ be any element in $\bar{A}\setminus A$.  Since $y\in \bar{A}$ there exists a sequence $(y_{n})\to y$ where each $y_{n}\in A$.  Since $h$ is continuous, $\lim_{n\to\infty} h(y_n)=h(y)$.  Similarly, $\lim_{n\to\infty}g(y_n)=g(y)$.  Since every $y_n$ is in $A$ it follows that $g(y_n)=h(y_n)$ for all $n$.  Hence, the limits of the sequences are the same.  I.e. $g(y)=h(y)$.  Since this is true for all $y\in \bar{A}\setminus A$ it follows that $g=h$.
A: Here is another, more general, proof that works in any Hausdorff space.
Proposition: Let $X$ be any topological space, and let $Y$ be a Hausdorff space. Suppose $A \subset X$ is dense in $X$, and that $f: A \to Y$ is a continuous map which extends continuously to some map $\bar{f}:X \to Y$. Then $\bar{f}$ is the unique extension of $f$ to $X$ which is continuous.
Proof: Suppose that the claim is false, so that the map $f$ admits two different continuous extensions $g \neq h$. Choose $x \in X$ such that $g(x) \neq h(x)$. Since $Y$ is Hausdorff, we can find disjoint neighborhoods $U$ and $V$ such that $g(x) \in U$ and $h(x) \in V$. Since $g$ and $h$ are continuous it follows that $g^{-1}(U) \cap h^{-1}(V)$ is open in $X$. Also $g^{-1}(U) \cap h^{-1}(V)$ is nonempty since it contains $x$. Since $A$ is assumed to be dense in $X$, we can deduce that $g^{-1}(U) \cap h^{-1}(V) \cap A$ is nonempty. Choose a point $a \in g^{-1}(U) \cap h^{-1}(V) \cap A$. Then $f(a)=g(a)=h(a) \in U \cap V$, which contradicts the fact that $U$ and $V$ are disjoint. $\Box$
Note: The above proposition is false if we do not assume that $Y$ is Hausdorff. For example, let $X$ be the real line with the usual topology. Let $A=\mathbb{R} \backslash \{0\}$, which is clearly dense in $X$. Let $Y$ be the real line with the cofinite topology (this is the topology where the closed sets are the finite sets). Then the map $f: X \to Y$ given by $x \mapsto |x|$ admits uncountably many different continuous extensions. Namely $\bar{f}(0)$ can be any real number.
