Proof verification, show that $f=g$. Let $X,Y$ be topological spaces such that Y a Hausdorff space. Let $A\subset X$ be with $\bar{A}=X$. Let $f,g:X\rightarrow Y$ be continuous functions such that $f(x)=g(x)$ for all $x\in A$. Show that $f=g$.
The closure of the $A$ is denoted by $\bar{A}$. The set of all limit points of $A$ is denoted by ${A}'$. 
My attempt:
We have to prove that $f(x)=g(x)$ for all $x\in X-A$. Suppose that there is a point $x\in X-A$ such that $f(x)\neq g(x)$. Since Y is a Hausdorff space, then $f(x)$ and $g(x)$ have disjoint neighborhoods $U$ and $V$, respectively. Since that $f$ and $g$ are continuous, there is a neighborhood $M$ of $x$ such that $f(M)\subset U$ and there is a neighborhood $N$ of $x$ such that $g(N)\subset V$. Because $U\cap V=\emptyset$, $f(M)\cap g(N)=\emptyset$.   
$X-A=\bar{A}-A=(A\cup {A}')-A={A}'-A$. Then every neighborhood of $x$ intersects $A$ in some point other than $x$ itself. This is particularly true for $M\cap N$, $$M\cap N\cap (A-\left \{ x \right \})=M\cap N\cap A\neq \emptyset.$$ Let $k$ be in $M\cap N\cap A$. Note that $$k\in M,N\implies f(k)\in f(M), g(k)\in g(N)\implies f(k)\neq g(k)$$ but $$k\in A\implies f(k)=g(k).$$ This is a contradiction, therefore $f=g$.
If I'm wrong give me a hint to be right, please. Thanks in advance.
 A: Your answer is fine.
For another way to answer this : In a Hausdorff space,the limit of a net is unique (standard exercise once you know what nets are, you should read up because they are interesting objects and sequences do not characterize closure always, hence the stronger notion of net). So if $x \in X$, there is a net $\{x_i\} \in A$ so that $x_i \to x$ as a net in the $X$ topology. By continuity (which preserves net convergence), $f(x_i) \to f(x)$, and $g(x_i) \to g(x)$, but $f(x_i) = g(x_i)$ for all $i$, so the same net is converging to two points $f(x),g(x)$, forcing $f(x)=g(x)$ for arbitrary $x \in X$.
A: Your proof is fine as is. You could make it "slick" as follows:
$Y$ is Hausdorff iff $\Delta_Y=\{(y,y): y \in Y\}$ is closed in $Y^2$.
$f,g: X \to Y$ continuous implies that $f \nabla g: X \to Y^2$ defined by
$(f \nabla g)(x) = (f(x),g(x))$ is continuous. 
And $f=g$ on $A$ implies $A \subseteq (f \nabla g)^{-1}[\Delta_Y]$ which is a closed subset so $X=\overline{A} = (f \nabla g)^{-1}[\Delta_Y]$ so $f=g$ on $X$.
The other answer (by Aston Villa etc.) with sequences can be made correct (for general spaces) by using nets instead and is also a quite natural approach IMO.
