I have two continuous function $f(x;\alpha)$ and $g(x;\alpha)$ that are continuous on $x \in [0,1]$ for any $\alpha \in [a,b]$.
Furthermore, $f(x;\alpha)$ and $g(x;\alpha)$ are both continuous on $\alpha$ for any $x$.
Suppose there is always a unique contact $x^{*}(\alpha)$ for any $\alpha$ such that $f(x^{*}(\alpha);\alpha)=g(x^{*}(\alpha);\alpha)$
How can I show that $x^{*}(\alpha)$ is continuous on $\alpha$?
It is not necessarily continuous.
Consider, for $0 < \alpha \le 1$, the piecewise-linear function $f(x;\alpha)$ whose graph consists of straight lines from $(0,1)$ to $(\alpha/2, 0)$ to $(\alpha, 1)$ to $(1,\alpha)$. For $\alpha = 0$ we take $f(x;0) = 1-x$. Note that $f(x;0) = \lim_{\alpha \to 0+} f(x;\alpha)$ for every $x \in [0,1]$, so this does satisfy the requirements of continuity in $x$ and in $\alpha$ (although it is not jointly continuous in $x$ and $\alpha$). Take $g(x;\alpha) = 0$. Then we have $x^*(\alpha) = \alpha/2$ for $\alpha > 0$, but $x^*(0) = 1$.
EDIT: To make it more explicit: for $\alpha > 0$, $$ f(x;\alpha) = \cases{1-\frac{2x}{\alpha} & for $0 \le x \le \alpha/2$\cr \frac{2x}{\alpha} - 1 & for $\alpha/2 \le x \le \alpha$\cr 1+\alpha-x & for $\alpha \le x \le 1$} $$