Find a function in the unit sphere of a function subspace with given properties

Preliminaries.

• $X$ is a compact Hausdorff space.
• $a$ is a fixed element of $X$.
• $C(X)$ denotes the space of real valued continuous functions on $X$
• $A=\{f \in C(X); f(a)=0\}$ is a Banach space respect to the following norm:$$\Vert f\Vert=\sup_{x,y \in X}\vert f(x)-f(y)\vert$$
• $\tilde f (x,y) \triangleq f(x)-f(y)$ and moreover
  1. $\tilde A=\{\tilde f; f\in A\}$ is Banach space respect to the usual supremum norm$$\Vert \tilde f\Vert=\sup_{x,y \in X} \vert \tilde f(x,y)\vert,$$
  2. $\tilde A $ is not a algebra,
  3. for any $x, y \in X$ with $x\neq y$, $\tilde A$ satisfies the following property: for any $ \varepsilon > 0 $ andopen neighborhoods $U, V$ of $x, y$ there exists a $\tilde f$ in the unit sphere of $\tilde A$ such that$$\tilde f(x,y)=1=\Vert \tilde f\Vert\;\text{ and }\;\vert\tilde f(z,t)\vert\leq\epsilon \quad \forall z,t \in U \cup V.$$

Problem:
Let $x_1,y_1$ be in $X$ and $\tilde f$ be in the unit sphere of $\tilde A$ such that $0\leq\tilde f(x_1,y_1)<1$: then there exists $\tilde h$ in the unit sphere of $\tilde A$ such that $\tilde h (x_1,y_1)=1$ and $\Vert\tilde h- \tilde f\Vert\leq 1-\tilde f(x_1,y_1)$

My proof.
By using Urysohn lemma we know that there exists a function $g$ in $C(X)$ such that $0\leq g\leq 1$ and $g(x_1)=1$, $g(y_1)=0$, $g(a)=0$ so $\tilde g$ is in the unit sphere of $\tilde A $ and $\tilde g(x_1,y_1)=1$.
Let's consider the function $h$ defined as:$$h=f+(1-\tilde f(x_1,y_1)) g.$$Clearly

1. $\tilde h$ is in $\tilde A$, and
2. $\tilde h(x_1,y_1)=1$ and $\Vert \tilde h-\tilde f\Vert <1-\tilde f(x_1,y_1)$ but $\tilde h$ is not in the unit sphere of $\tilde A$

I think that if for any $0<r<1$ we can find a function $ \tilde g_r$ in the unit sphere of $\tilde A$ such that $\tilde g_r(x_1,y_1)=1$, then the function $\tilde h_r= r\tilde f+(1-r\tilde f(x_1,y_1))\tilde g_r$ is the desired one, but I'm unsure on how to do this.