Let f and g continuous, then if x contained in open ball with center a, and f(x)=g(x), then f(a)=g(a) Let ($X, d$), ($Y, d'$) be two metric spaces, $ a \in X$, and $f, g : X \rightarrow Y$ two functions.
Suppose $f$ and $g$ continuous. Prove that if any open ball centered on $a$ contains a point $x$ such that $f(x) = g(x)$, then $ f(a) = g(a)$.
I am not sure how to approach this problem, but my intuition tells me that either $x$ is arbitrary close to $a$ (meaning that $a=x$), or $f = g$ at least where $x\in B(a,r)$, such that $f(B(a,r)) = g(B(a,r))$.
I have tried defining another ball $B$ with center on $x$ and same radius $r$, such that $a\in B(x,r)$. Because $f$ and $g$ are continuous, then $\exists \delta_1,\delta_2$ such that $f(B)\subset B(f(x),\delta_1)$ and $g(B) \subset B(g(x),\delta_2)$. Then $B(f(x),\delta_1)$ and $B(g(x),\delta_2)$ are concentric, meaning that they have the same center, $f(x)=g(x)$.
Now take the smallest of $\delta_1$or $\delta_2:\delta = min\{\delta_1,\delta_2\}$. If we make $\delta$ arbitrary small, then in order for $f(a)$ to be kept inside any of the balls, then we have to make $a$ arbitrary close to $x$. Therefore $a=x$ and $f(a)=f(x)=g(x)=g(a)$.
I am not sure of this proof, so I would assistance in solving this problem.
Thanks in advance!

For each natural $n$, let $x_n\in B\left(a,\frac1n\right)$ such that $f(x_n)=g(x_n)$. Then $d(x_n,a)<\frac1n$. So, $\lim_{n\to\infty}x_n=a$ and, since $f$ and $g$ are continuous,$$f(a)=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}g(x_n)=g(a).$$
