times X,
\end{equation*}
i.e.~$\tilde{\phi}|_{W}=\phi|_{W}$. The existence of $\tilde{\phi}$ compact is guaranteed by an application of Urysohn embedding theorem 
(see, for instance, \cite[Theorem~B.11, p.~597]{GrDu-2003}) 
and Dugundji extension theorem (cf.~\cite{Du-1951} and \cite[Theorem~7.4, p.~163--164]{GrDu-2003}) together with the results in \cite[\textsection~6.2, p.~116--118]{GrDu-2003}.

Let $r\colon X \to A$ be a retraction. We introduce the operator
\begin{equation*}
\psi=(\psi_{1},\psi_{2})\colon \mathbb{R}\times X\to\mathbb{R}\times A, 
\end{equation*}
defined as follows
\begin{equation*}
\psi_{1}(t,x):=\tilde{\phi}_{1}(t,x), \qquad \psi_{2}(t,x):=r(\tilde{\phi}_{2}(t,x)).
\end{equation*}
We observe that $\psi$ is a compact map (by the compactness of $\tilde{\phi}$ and the continuity of $r$). We define
\begin{equation*}
\hat{A}:=\overline{\psi_{2}(\mathbb{R}\times X)}.
\end{equation*}
The set $\hat{A}$ is nonempty, compact and contained in $A$ (since $A$ is closed, by \cite[Theorem~5.1, p.~18]{Hu-1965}).

We also note that if $\tilde{z}=(\tilde{t},\tilde{x})$ is a fixed point of $\psi$ with $\tilde{z}\in W$ and $\phi_{2}(\tilde{z})\in A$, then
\begin{equation*}
\tilde{t}=\psi_{1}(\tilde{z})=\tilde{\phi}_{1}(\tilde{z})=\phi_{1}(\tilde{z})
\end{equation*}
and
\begin{equation*}
\tilde{x}=\psi_{2}(\tilde{z})=r(\tilde{\phi}_{2}(\tilde{z}))=r(\phi_{2}(\tilde{z}))=\phi_{2}(\tilde{z}),
\end{equation*}
hence $\tilde{z}\in W$ is a fixed point of $\phi$.

Now we study the fixed point problem
\begin{equation*}
x=\psi_{2}(t,x), \quad x\in X,
\end{equation*}
where $t$ is treated as a parameter in $\mathopen{[}a,b\mathclose{]}$.
From the definition of $\hat{A}$, it is obvious that $\psi_{2}(t,x)\in \hat{A}\subseteq A$ for all
$(t,x)\in\mathopen{[}a,b\mathclose{]}\times X$. We fix $r>0$ such that the open ball $B(0,r)$ contains the closed and bounded set $\hat{A}$. It
follows that, for all $t\in\mathopen{[}a,b\mathclose{]}$,
\begin{equation*}
x-\psi_{2}(t,x)\neq0, \quad  \forall \, x\in\partial B(0,r).
\end{equation*}
Consequently, the fixed point index $i_{X}(\psi_{2}(t,\cdot),B(0,r))$ is well defined and constant, for $t\in\mathopen{[}a,b\mathclose{]}$.
Consider now the compact homotopy $h\colon\mathopen{[}0,1\mathclose{]}\times B[0,r]\to X$, $h_{\lambda}(x):=\lambda\psi_{2}(t,x)$. 
For every $t\in\mathopen{[}a,b\mathclose{]}$ we have that
\begin{equation*}
x-\lambda\psi_{2}(t,x)\neq0, \quad \forall \, \lambda\in\mathopen{[}0,1\mathclose{]}, \; \forall \, x\in\partial B(0,r),
\end{equation*}
since $\lambda\psi_{2}(t,x)\in\lambda\hat{A}\subseteq B(0,r)$ for every $\lambda\in\mathopen{[}0,1\mathclose{]}$ and $x\in X$.
 
By the homotopy invariance property, the integer $i_{X}(h_{\lambda},B(0,r))$ is constant with respect to $\lambda\in\mathopen{[}0,1\mathclose{]}$.
In particular, for every $t\in\mathopen{[}a,b\mathclose{]}$,
\begin{equation*}
i_{X}(\psi_{2}(t,\cdot),B(0,r))=i_{X}(0,B(0,r))=1.
\end{equation*}

Therefore we can apply the \textit{Leray-Schauder continuation principle} (Theorem~\ref{LScont_prin}), which ensures that the nonempty set
\begin{equation*}
\Sigma:=\bigl{\{}(t,x)\in\mathopen{[}a,b\mathclose{]}\times B(0,r) \colon x=\psi_{2}(t,x)\bigr{\}}
\end{equation*}
contains a compact and connected (\textit{continuum}) set $S$ such that 
\begin{equation*}
S\cap \mathcal{C}_{l}\neq\emptyset \quad \text{ and } \quad S\cap \mathcal{C}_{r}\neq\emptyset.
\end{equation*}
As consequence, $p_{1}(S)=\mathopen{[}a,b\mathclose{]}$, where $p_{1}\colon\mathbb{R}\times X\to \mathbb{R}$, $p_{1}(t,x)=t$, 
and for every open $\tilde{A}$ containing $\hat{A}$
\begin{equation*}
p_{2}(S)\subseteq \hat{A}\subseteq \tilde{A},
\end{equation*}
where $p_{2}\colon\mathbb{R}\times X\to X$, $p_{2}(t,x)=x$.

Fix an open and bounded set $\tilde{A}$ such that $\hat{A}\subseteq \tilde{A}$ (for example $\tilde{A}=B(0,r)$, as before) 
and fix $\varepsilon\in\mathbb{R}$ such that $0<\varepsilon<\text{dist}(\hat{A}, X\setminus \tilde{A})<+\infty$. By the continuity of the retraction $r$,
for every $x\in X$ we can find a positive number $\delta_{x}=\delta_{x}(\varepsilon)$, with $0<\delta_{x}<\varepsilon$, such that for every $y\in X$ with
$\|y-x\|<\delta_{x}$ it holds that $\|r(y)-r(x)\|<\varepsilon$.

Consider the open covering of $S$ defined as
\begin{equation*}
\bigl{\{}\,\mathopen{]}t-\varepsilon,t+\varepsilon\mathclose{[}\times B(x,\delta_{x})\colon (t,x)\in S \bigr{\}}.
\end{equation*}
Since $S$ is compact, we can extract a finite sub-covering
\begin{equation*}
\bigl{\{}\,\mathopen{]}t_{i}-\varepsilon,t_{i}+\varepsilon\mathclose{[}\times B(x_{i},\delta_{x_{i}})\colon i=1,\ldots,m \bigr{\}},
\end{equation*}
where $(t_{i},x_{i})\in S$, so $t_{i}\in\mathopen{[}a,b\mathclose{]}$ and $x_{i}\in \hat{A} \subseteq A$, for all $i\in\{1,\ldots,m\}$.

Now we set
\begin{equation*}
\mathcal{U}_{\varepsilon}
:=\bigcup_{i=1}^{m} \, \mathopen{]}t_{i}-\varepsilon,t_{i}+\varepsilon\mathclose{[}\times B(x_{i},\delta_{x_{i}}),
\end{equation*}
which is contained in $\mathopen{]}a-\varepsilon,b+\varepsilon\mathclose{[}\times \tilde{A}$, because
$0<\delta_{x_{i}}<\varepsilon<\text{dist}(\hat{A}, X\setminus \tilde{A})$, for all $i=1,\ldots,m$.
The set $\mathcal{U}_{\varepsilon}$ is open and connected (because $S\subseteq\mathcal{U}_{\varepsilon}$ 
and $(t_{i},x_{i})\in S\cap (\mathopen{]}t_{i}-\varepsilon,t_{i}+\varepsilon\mathclose{[}\times B(x_{i},\delta_{x_{i}}))\neq\emptyset$ for all
$i\in\{1,\ldots,m\}$).