Let $\Gamma: \mathcal{Y} \twoheadrightarrow \mathcal{X}$ be an upper hemicontinuous correspondence with non-empty values and a closed graph. A continuous selection function in general does not exist, but is it always possible to construct a finite number of continuous functions, $f\_i: \mathcal{Y} \to \mathcal{X}, i=\{1,...,K\}$, such that $(\cup\_{i=1}^K f\_i(y)) \ \cap \ \Gamma(y) \neq \emptyset $?



Collection of continuous selections for upper hemicontinuous map
$\newcommand{\Ga}{\Gamma}\newcommand{\N}{\mathbb N}$Not in general.


E.g., let $Y:=\mathcal Y:=[0,1]$, $X:=\mathcal X:=[-1,1]$, $\Ga(0):=[-1,1]$, and $\Ga(y):=\{g(y)\}$ for $y\in(0,1]$, where $g(y):=\sin\frac1y$.


Then $\Ga\colon Y\twoheadrightarrow X$ is an [upper hemicontinuous](https://en.wikipedia.org/wiki/Hemicontinuity) correspondence with non-empty values and a closed graph.


Suppose now that, for some natural $K$, there is a family $(f\_i)\_{i\in[K]}$ of continuous functions from $Y$ to $X$ such that $\bigcup\_{i\in[K]}\{f\_i(y)\}\cap\Ga(y)\ne\emptyset$ for all $y\in Y$. Here, as usual, $[n]:=\{1,\dots,n\}$. Then for each $y\in(0,1]$ there is some $I(y)\in[K]$ such that
\begin{equation\*}
 f\_{I(y)}(y)=g(y). \tag{1}
\end{equation\*}


Take any natural $L>K$ and let $a\_1,\dots,a\_L$ be any pairwise distinct points in $X=[-1,1]$. Then for each $j\in[L]$ there is a sequence $(y\_{j,1},y\_{j,2},\dots)$ in $(0,1]$ such that $y\_{j,k}\to0$ and
$$g(y\_{j,k})\to a\_j$$
as $k\to\infty$.
For each $j\in[L]$, $\bigcup\_{i\in[K]}R\_{i,j}=\N$, where $R\_{i,j}:=\{k\in\N\colon I(y\_{j,k})=i\}$, where $I(\cdot)$ is as in (1). So,
for each $j\in[L]$ there is some $i\_j\in[K]$ such that the set
\begin{equation\*}
 Q\_j:=R\_{i\_j,j}=\{k\in\N\colon I(y\_{j,k})=i\_j\}
\end{equation\*}
is infinite. Since $L>K$, the map $[L]\ni j\mapsto i\_j\in[K]$ cannot be injective (the pigeonhole principle). So, $i\_{j\_1}=i\_{j\_2}=:i\_\*\in[K]$ for some distinct $j\_1,j\_2$ in $[L]$. So, for each $r\in\{1,2\}$ and all $k\in Q\_{j\_r}$
\begin{equation\*}
 f\_{i\_\*}(y\_{j\_r,k})=f\_{i\_{j\_r}}(y\_{j\_r,k})=f\_{I(y\_{j\_r,k})}(y\_{j\_r,k})=g(y\_{j\_r,k})\to a\_{j\_r}
\end{equation\*}
as $Q\_{j\_r}\ni k\to\infty$, and also $y\_{j\_r,k}\to0$ as $k\to\infty$.


Therefore and because $a\_{j\_1}\ne a\_{j\_2}$, the function $f\_{i\_\*}$ is not continuous at $0\in Y$, a contradiction. $\quad\Box$


*Remark:* The proof would be much simplified if, in accordance with the mentions of "continuous selection" in the title and body of the posted question, we assumed in addition that the graphs of the functions $f\_i$ are contained in the graph of $\Ga$.


