Let $X$ be a topological space and let $(x_i)_i \in I$ be a family of elements of $X$. If $D$ is an ultrafilter on $I$ and $x \in X$, we write $$x_i \mapsto_{D} x$$ if for every neighborhood $U$ of $x$, the set $\{ i \in I : x_i \in U \}$ is in the ultrafilter $D$.
I want to show that if $(a_i) \mapsto_{D} a$ and $f_i \mapsto_{D} f$ then $ Sup_{x}f_i(x, a_i) \mapsto_{D} Sup_{x} f(x, a)$, where $f_i : \mathbb{R}^2 \rightarrow [0,1]$ is continuous.
Example. It seems $X = \mathbb R$ in your statement...
Let $I = \{1,2,3,\dots\}$. Define $$ a_i = \frac{1}{i},\qquad a=0 . $$ So of course $a_i \to a$.
Define $f_i$ by: $f_i(x,a_i) = 1$, $f_i(x,y) = 0$ for $y \le a_{i+1}$ or $y \ge a_{i-1}$ and interpolate linearly on each side of $a_i$. Then for each $i$, we have $f_i(x,y)$ is independent of $x$ and continuous in $y$, therefore jointly continuous. Define $f(x,y) = 0$, also jointly continuous. But $$ f_i(x,y) \to 0,\qquad\text{for all } x,y, $$ so $f_i \to f$. [Of course this convergence is not uniform, but the problem did not require that.]
Next, $$ \lim_i \sup_x f_i(x,a_i) = 1 \ne 0 = \lim_i \sup_x f_i(x,a) $$
but this is just because $$ f_i(x,a_i) = 1 \text{ for all } x,i, \\ f_i(x,a) = 0 \text{ for all } x,i . $$