Show that $ h(x)=\max \{f(x,y) : y \in [0,1] \} $ is continuous Let $ f(x,y) \ $ be a continuous real-valued function on the unit square $ [0,1] \times [0,1]$. 
Show that $$ h(x)=\max \{\,f(x,y) : y \in [0,1] \},  $$  is also continuous.
Answer.
Since   $ f(x,y)$ is continuous, then $ \max \{f(x,y) \}$ is also continuous on $[0,1 ] \times [0,1]$. 
Thus for any fixed values of $ y \in [0,1] \ $  ,  $ \max \{f(x,y) \}$  is also continuous .
i.e.,  $ \max \{f(x,y): y \in [0,1] \}=h(x) \ $ is also  continuous. 
But I need  a $ \varepsilon-\delta \ $ proof. Is there any?
 A: Since $f$ is continuous, then for every $\varepsilon>0$, there exists a $\delta>0$, such that
$$
|x_1-x_2|+|y_1-y_2|<\delta\quad\Longrightarrow\quad |\,f(x_1,y_1)-f(x_2,y_2)|<\varepsilon
$$ 
Consider $x_1,x_2\in[0,1]$, with $|x_1-x_2|<\delta$. Then there exists $y_1,y_2\in [0,1]$, such that
$$
|h(x_1)-h(x_2)|=
\Big|\max_y f(x_1,y)-\max_y f(x_2,y)\Big|=
| f(x_1,y_1)-f(x_2,y_2)|.
$$
Clearly, 
$0 \le f(x_1,y_1)-f(x_2,y_1)<\varepsilon\quad$ and
$\quad 0\le f(x_2,y_2)-f(x_1,y_2)<\varepsilon$.
Thus,
$$
f(x_1,y_1)<f(x_2,y_1)+\varepsilon\le f(x_2,y_2)+\varepsilon \tag{1}
$$
and
$$
f(x_2,y_2)<f(x_1,y_2)+\varepsilon\le f(x_1,y_1)+\varepsilon \tag{2}
$$
Combining (1) and (2), we obtain
$$
-\varepsilon< f(x_1,y_1)-f(x_2,y_2)<\varepsilon
$$
or equivalently
$$
|h(x_1)-h(x_2)|<\varepsilon.
$$
A: Suppose that $h$ is not continuous at some point $x_0.$ then there exists a number $\varepsilon >0$ and a sequence $x_n \to 0 $ such that $$|h(x_n ) -h(x_0 )|\geq \varepsilon.$$ Since the functions $\xi_n (y) =f(x_n ,y ) $ are continuous on $[0,1],$ thus for any $n$ there exists $y_n $ such that $h(x_n )= \xi_n (y_n) =f(x_n , y_n ).$ But the interval $[0,1] $ is compact therefore $y_{n_j} \to y_0 \in [0,1],$ for some sequence $(n_j ).$ Hence $$f(x_{n_k} , y^{\chi} )-f(x_0 ,y^{\chi})\leq f(x_{n_k} , y_{n_k} )-f(x_0 ,y^{\chi})\leq -\varepsilon \vee f(x_{n_k} , y^{\chi} )-f(x_0 ,y^{\chi})\geq f(x_{n_k} , y_{n_k} )-f(x_0 ,y^{\chi})\geq \varepsilon$$
where $y^{\chi}$ is such that $h(x_0 ) = f(x_0 ,y^{\chi} ) .$
But the last implies that $$0=f(x_0 , y^{\chi} )-f(x_0 ,y^{\chi})\leq f(x_0 , y_0 )-f(x_0 ,y^{\chi})\leq -\varepsilon \vee 0=f(x_0 , y^{\chi} )-f(x_0 ,y^{\chi})\geq f(x_0 , y_0 )-f(x_0 ,y^{\chi})\geq \varepsilon$$
Contradiction
A: You need to use this inequality.
$$|max \ f(x) - max \ g(x)| \le max|f(x) - g(x)|$$
Then, write the difference below.
$$|h(x)-h(x_0)|=|max\{f(x,y):y\in [0,1]\}-max\{f(x_0,y):y\in [0,1]\}|$$
Note that the two maximum functions, in the expression above, can be considered as $f(x)$ and $g(x)$ that we used in the inequality. So, using the inequality, the expression becomes
$$|max\{f(x,y):y\in [0,1]\}-max\{f(x_0,y):y\in [0,1]\}| \leq max\{|f(x,y)-f(x_0,y)|:y\in [0,1]\}$$
then, we use the continuity of $f(x,y)$ to say
$$\forall\epsilon \hspace{1mm}\exists \hspace{1mm} \delta|\hspace{1mm} 
|d(x-x_0,y-y)|=|d(x-x_0,0)|=|x-x_0|<\delta \implies |f(x,y)-f(x_0,y)|<\epsilon$$
Now, you need to choose the delta to be as below, knowing that for a data point $(x,y)$, $\delta$ is a function of $\epsilon$. So write $\delta(\epsilon,x,y)$
$$\delta(\epsilon) = inf\{\delta(\epsilon,x,y)|x,y\}$$
with this delta, you can go in the inverse direction.
