Showing a function is continuous(real analysis) 
Hello, I'm struggling showing that this function g(x) is continuous.  I've proved before that the composition of two questions, g o f is continuous and this question slightly reminds me of that.  I was thinking of maybe showing that preimages of open sets in g(x) are open sets or should I instead use the fact that a function is continuous that if pn converges to p then g(pn) converges to g(p)? Or I could use the epsilon delta definition of continuity?  I could really use any help, guidance, and suggestions?  Thanks.
 A: It suffices to show that the preimage of any open interval $(a,b)$ under $g$ is open. This is equivalent to showing that the preimages of both $(a,\infty)$ and $(-\infty,b)$ are open.
Now if $g(x)>a$, then we must have $f(x,y)>a$ for some $y$.  From here it is easy to show that $g(x')>a$ for all $x'$ in a neighborhood of $x$.
Now if $g(x)<b$, then you must have $f(x,y)<b$ for all $y\in [0,1]$. So for each $y\in [0,1]$there exists an open neighborhood $U_y$ of $x$ and an open neighborhood $V_y$ of $y$ such that $f(x',y')$ for all $x'\in U_y$ and all $y'\in V_y$. By compactness of $[0,1]$, finitely many $V_y$ cover $[0,1]$. Let $U$ be the finite intersection of the corresponding $U_y$. Show that $f(x',y)<b$ for all $x'\in U$ and all $y\in[0,1]$. 

Here is an alternative proof of the second part using sequences. Suppose for the sake of contradiction that $g(x)<b$, but that there is no neighborhood of $x$ such that $g(x')<b$ for all $x'$ in the neighborhood. Then there is for all $n$ some $x_n\in B_{1/n}(x)$ and some $y_n$ such that $f(x_n,y_n)\geq b$. There is a subsequence $\langle y_{n_m}\rangle$ of the sequence $\langle y_n\rangle$ such that $\langle y_{n_m}\rangle$ converges to some $y\in [0,1]$. Use continuity of $f$ to show that $f(x,y)\geq b$ and obtain a contradiction.
A: I would suggest using the $\epsilon-\delta$ definition of continuity.  You will also want to use the fact that $[0,1]\times[0,1]$ is compact, so $f$ is not just continuous but uniformly continuous.  As a starting point, using uniform continuity of $f$, can you prove that for any $\epsilon>0$ there exists a $\delta>0$ such that $g(x')\geq g(x)-\epsilon$ whenever $|x'-x|<\delta$?
A stronger hint is hidden below:

 Let $y$ be such that $f(x,y)=g(x)$.  What does the uniform continuity of $f$ then tell you when you apply it at the point $(x,y)$?

