Let us show how to find such a retraction for $n=2$ (the general case is analogous but technically more complicatedI do not know if this method generalizes to higher dimensions).
Without loss of generalityReplacing the triangulation by a finer triangulation, we can assume that for each triangle $T$ with $T\not\subseteq C$, one vertex of the triangulation either$T$ does not belong to $C$ meets the.
How to find such a triangulations? Assuming that $T\not\subseteq C$, we can find an interior point $v$ of $T$ that does not belong to $C$ and replace the triangle $T$ by 3 subtriangles having $v$ as a vertex.
Also we can assume that either $T\subseteq\mathbb R^2\setminus C$ or else$T$ has a vertex in $C$ intersects at most two sides of the triangle. IfAssuming that $T$ has no vertices in $C$ intersects three sides of the triangle but does not intersect the interior$T\cap C\ne\emptyset$, then we can divide this triangle intochoose a point $c\in T\cap C$ and replace the triange $T$ by two or three triangles having at most two sides intersecting $C$$c$ as a vartex.
LetTherefore, we lose no generality assuming that each triangle $K^{(0)}$ be the set of vertices$T$ of the triangulation and lethas one of the following properties:
$T\subseteq C$;
$T\cap C=\emptyset$;
$T$ has one vertex in $C$ and one vertex outside of $C$;
If two vertices $u,v$ of $T$ do not belong to $C$, then the side $[u,v]$ does not intersect $C$.
A triangle $K$$T$ will be the family of subsetscalled $\sigma\subset K^{(0)}$ whose convex combinations are verticesdifficult if it has one vertex say $u$ outside $C$, edges or triangles oftwo vertices $v,w$ in $C$ and the triangulationside $[v,w]$ is not a subset of $C$. In this case choose any point $c[v,w]\in [v,w]\setminus C$. The setpoints $K$$c[v,w]$ can be written aschosen so that for two difficult triangle sharing the union $K=\bigcup_{n=1}^3K^{(n)}$ wherecommon side $K^{(n)}=\{\sigma\in K:\sigma$ has exactly$[v,w]$ the point $n+1$ elements$\}$$c[v,w]$ is the same.
ForNow for every triangle $T$ of the triangulation we define a setfunction $\sigma$$r_T\setminus C:T\to T\setminus C$ such that $r_T\circ r_T=r_T$ as follows. In case (1), let $\bar\sigma$$r_T$ be the convex hull ofempty map and in case $\sigma$,(2) $\partial\sigma$$r_T$ be the combinatorial boundaryidenity map of $\bar\sigma$$T$. In the remaining cases, the triangle $T$ has one vertex in $C$ and one vertex outside of $\sigma^\circ=\bar\sigma\setminus\partial\sigma$$C$. If the triangle $T$ is not difficult, then it has two vertices $u,v$ such that the side $[u,v]$ either is contained in $C$ or is disjoint with $C$. If $[u,v]$ is contained in $C$, then let $r_T:T\setminus C\to\{w\}$ be the combinatorial interiorconstant map into the unique vertex $w\notin C$ of $\bar\sigma$$T$.
LetIf $K_C=\{\sigma\in K:\bar\sigma\cap C\ne\emptyset\}$$[u,v]\cap C=\emptyset$, then the third vertex $w$ of $T$ belongs to $C$ and we can apply the Urysohn lemma to find a function $V=\bigcup\{\sigma^\circ:\sigma\in K_C\}$$r_T:T\setminus C\to[u,v]$ such that $r_T[[w,u]\setminus C]=\{u\}$, $r_T[[w,v]\setminus C]=\{v\}$, and $r_T(x)=x$ for every $x\in [u,v]$.
Then $V$ is an open neighborhoodIt remains to consider the case of a difficult triangle $C$$T$. We claim that there existsSuch a retraction oftriangle has one vertex $\mathbb R^2\setminus C$ onto$u$ outside of $\mathbb R^2\setminus V$. For every simplex$C$, two vertices $\sigma\in K^{(2)}\cap K_C$ the set$v,w$ in $C$ either intesects $\sigma^\circ$ of intersect at most two 1-simplexes ofand the boundarypoint $\partial \sigma$$c[v,w]\in [v,w]\setminus C$. In bothTwo cases we can construct a retraction $r_\sigma:\bar\sigma\setminus C\to\partial \sigma\setminus C$. For every $\sigma\in K^{(2)}\setminus K_C$ let $r_\sigma$ be the identity map of $\bar\sigma$are possible.
There exists a path $\gamma:[0,1]\to T\setminus C$ such that $\gamma(0)=u$ and $\gamma(1)=c[v,w]$. We can assume that $\gamma$ is injective and hence its image $A_T=\gamma[0,1]$ is an arc with endpoints $u$ and $c[v,w]$. Using the Urysohn Lemma, we can find a continuous function $r_T:T\setminus C\to A_T$ such that $r_T[([u,v]\cup[u,w])\setminus C]\subseteq \{u\}$, $r_T[[v,w]\setminus C]\subseteq\{c[v,w]\}$ and $r_T(a)=a$ for every $a\in A_T$.
No such a path $\gamma$ exists. Then the points $u$ and $c[v,w]$ belong to distinct connected components of $T\setminus C$. In this case we can choose a continuous map $r_T:T\setminus C\to\{u,c[v,w]\}$ such that $r_T[([u,v]\cup[u,w])\setminus C]\subseteq\{u\}$ and $r_T[[v,w]\setminus C]\subset\{c[v,w]\}$.
The uniondefinitions of the retractionmaps $r_\sigma$$r_T$ ensure that they agree on the intersections of their domains. Consequently, the union $\sigma\in K^{(2)}$$r=\bigcup_T r_T$ of these maps is a continuous function $r:\mathbb R^2\setminus C\to\mathbb R^2\setminus C$ such that $r\circ r=r$. So, $r$ is a retraction of $\mathbb R^2\setminus C$ onto the setclosed subset $S=(\mathbb R^2\setminus V)\cup\bigcup_{\sigma\in K^{(1)}\cap K_C}(\bar\sigma\setminus C)$. Then do$F$ which can be written as the same withunion of the set $K^{(1)}\cap K_C$: for every 1-simplextriangles of the triangulation that do not intersect $\sigma\in K^{(1)}\cap K_C$$C$, usesome vertices of the facttriangles that $\bar\sigma\cap C\ne\emptyset$ and find a retractionintersect $r_\sigma:\bar\sigma\setminus C\to\partial \sigma\setminus C$. The union of those retractions$C$ and the identity maps $r_\sigma$ forarcs $\sigma\in K^{(2)}\setminus K_C$ retract the set$A_T$ of difficult triangles $S$ onto(of the set $$S'=(\mathbb R^2\setminus V)\cup \bigcup_{\sigma\in K^{(2)}\cap K_C}(\sigma^{(0)}\setminus C).$$first type).
SinceThe choice of the settriangulation $S'\setminus(\mathbb R^2\setminus V)$$T$ (as sufficiently fine) implies that $V=\mathbb R^2\setminus F$ is finite, here exists a retrationneighborhood of $S'$ onto$C$ with $\mathbb R^2\setminus V$$\bar V\subset U$. So, our final retractionThen $\mathbb R^2\setminus C\to\mathbb R^2\setminus V$$r{\restriction}U\setminus C$ is the compositionrequired retraction of the retractions $\mathbb R^2\setminus C\to S\to S'\to\mathbb R^2\setminus V$$U\setminus C$ onto $U\setminus V$.