It's a good idea to use the Baire Category Theorem, but there is still work to be done to verify that the sets are open and dense.

Here's a summary of the proof for fast readers (for more clarification, I provide all the gory details afterward):

The set of homotopies avoiding $x_n$ is clearly open. Let $h:\mathbb{D}\rightarrow \mathbb{R}^d$ be a homotopy from $\gamma$ to $y$ and suppose $x_n\not\in \gamma(S^1).$ Approximate $h$ by a piecewise-linear (PWL) map $h^*,$ and let $g^t$ be the straight-line homotopy from $h$ to $h^*.$ Let $\rho:\mathbb{D}\rightarrow[0,1]$ be a continuous map with $\rho|_{\partial\mathbb{D}}=0$ and $\rho|_U=1$ where $U\subset \textrm{int}(\mathbb{D})$ is an open neighborhood of $h^{-1}(x_n).$ Defining $$H = g^{\rho(-)}(-),$$ we then have that $H$ is close to $h,$ $H|_{\partial \mathbb{D}}=\gamma,$ and $H|_U$ is PWL. Moreover, we can take $U$ sufficiently large so that $H(\mathbb{D}\setminus U)$ is sufficiently close to $\gamma(S^1),$ so that $x_n\not\in H(\mathbb{D}\setminus U)$. Because $H|_U$ is PWL, $H(U)$ has no interior, so it follows that $x_n$ is not in the interior of $H(\mathbb{D}).$ Hence we can find a point $y\not\in H(\mathbb{D})$ that is close to $x_n.$ In particular $B_\varepsilon(y) \cap H(\mathbb{D})=\emptyset.$ Then, apply a perturbation $f_y:\mathbb{R}^d\rightarrow \mathbb{R}^d,$ that enlarges the neighborhood $B_\varepsilon(y)$ so that $x_n\in f_y(B_\varepsilon(y)).$ We can certainly choose $f_y$ to be small and equal to identity on $\gamma(S^1).$ Hence the perturbed homotopy $f_y\circ H$ is close to $h,$ avoids $x_n,$ and $(f_y\circ H)|_{\partial \mathbb{D}} = \gamma$ as desired.

**Formal proof:**

Let $\{x_1,x_2,\dotsc\}\subset \mathbb{R}^d$ be given, and let $\gamma:S^1\rightarrow \mathbb{R}^d\setminus \{x_1,x_2,\dotsc\}$ be a loop.

Denote $\mathbb{D} = \{z\in\mathbb{C}\mid \lvert z\rvert\leq 1\}$, and identify $\partial \mathbb{D} = S^1$, and let $C(\mathbb{D},\mathbb{R}^d)$ be the space of continuous functions from $\mathbb{D}$ to $\mathbb{R}^d$ equipped with the uniform norm. Denote the subset
$$\mathcal{H} = \{h\in C(\mathbb{D},\mathbb{R}^d)\mid h\rvert_{\partial \mathbb{D}} = \gamma\}.$$
Thus, $\mathcal{H}$ can be viewed as the space of all homotopies of $\gamma$ to a point.
The uniform limit theorem ensures that $\mathcal{H}$ is complete. In particular, $\mathcal{H}$ is a Baire-space.

For each $x_n$, define $$S_n = \{h\in \mathcal{H}\mid x_n\notin h(\mathbb{D})\}.$$

If $h_m$ is a sequence in $\mathcal{H}\setminus S_n$, then there is a sequence $z_m\in \mathbb{D}$ for which $h_m(z_m)=x_n$. And, if $h_m\rightarrow h\in \mathcal{H}$, then—because convergence is uniform—we have that for every $\varepsilon$ there is an $m$ large enough so that $$d(h(z_m),x_n) = d(h(z_m),h_{m}(z_{m}))<\varepsilon.$$ So for any convergent subsequence $z_{m_j}\rightarrow z$ we have that $$h(z) = h(\lim_{j\rightarrow\infty}z_{m_j})= \lim_{j\rightarrow\infty}h(z_{m_j}) = x_n.$$ In particular, $h\notin S_n$. Thus $H\setminus S_n$ is closed, and so $S_n$ is open.

To see that $S_n$ is dense in $\mathcal{H}$, let $h\in \mathcal{H}\setminus S_n$ and $\varepsilon>0$ be given, and suppose that $\varepsilon$ is sufficiently small so that the open ball $B_{3\varepsilon}(x_n)$ is disjoint from $\gamma(S^1)$ (this can be done because $\gamma(S^1)$ is closed and disjoint from every $x_n$).

For every $m\in\mathbb{N}$, let $T_m$ be a triangulation of $\mathbb{D}$ so that every triangle $\tau\in T_m$ has diameter $\lvert\tau\rvert<2^{-m}$. Let $h_m$ be the piecewise linear (PWL) approximation of $h$; that is, $h_m$ is linear on every triangle $\tau\in T_m$, and $h_m$ agrees with $h$ on every vertex of every triangle.

Then, because $h$ is uniformly continuous, and the vertices of the triangles in $T_m$ are $2^{-m}$ dense in $\mathbb{D}$, we can certainly take an $m\in \mathbb{N}$ sufficiently large so that

$$\lVert h_m - h\rVert_{\infty}\leq\varepsilon.$$

If $h_m\in S_n$, we are done. So suppose otherwise that $x_n\in h_m(\mathbb{D})$.

As $h$ is uniformly continuous, we can take $0<R<1$ sufficiently large so that $h(\mathbb{D}\setminus B_R(0))\subset B_\varepsilon(\gamma(S^1)).$ Also, let $R$ be large enough so that $h^{-1}(x_n)\subset B_R(0).$

Let $g^t$ be the straight-line homotopy from $h$ to $h_m,$ and let $\rho:\mathbb{D}\rightarrow[0,1]$ be a continuous map with $\rho|_{\partial\mathbb{D}}=0$ and $\rho|_{B_R(0)}=1.$ Defining $$H = g^{\rho(-)}(-),$$ we then have that

$$\begin{aligned}||H-h_m||_\infty &= ||(1-\rho)h + \rho h_m - h_m||_\infty \\ &\leq ||1-\rho||_\infty\cdot||h-h_m||_\infty \\&\leq \varepsilon. \end{aligned}$$

Also, for every $z\in \mathbb{D}\setminus B_R(0),$ we have $$\begin{aligned}\inf_{s\in[0,1]}|H(z)-\gamma(s)| &= |H(z) - h_m(z)+h_m(z)-h(z)+h(z)-\gamma(s)| \\ &\leq \inf_{s\in[0,1]}|h(z)-\gamma(s)| + 2\varepsilon \\
&< 3\varepsilon,\end{aligned}$$ and because $B_{3\varepsilon}(x_n)\cap \gamma(S^1)=\emptyset,$ this shows that $x_n\not \in H(\mathbb{D}\setminus B_R(0)).$

Now, consider that $H(B_R(0)) = \bigcup_{\tau\in T_m}h_m(\tau\cap B_R(0))$, and each $h_m(\tau\cap B_R(0))$ has no interior in $\mathbb{R}^d$ (this is where we need $d\geq 3$). Also, $T_m$ contains finitely many triangles, and so $H(B_R(0))$ is a finite union of sets with no interior, and hence has no interior.

Because $x_n\not\in H(\mathbb{D}\setminus B_R(0)),$ and $H(B_R(0))$ has no interior, we can find a $y\not\in H(\mathbb{D})$ for which $\lvert x_n-y\rvert<\varepsilon$.

Next, define

$$f_{y}:\mathbb{R}^d\setminus\{y\}\rightarrow\mathbb{R}^d: f_{y}(v) = (\lvert v-y\rvert+\varepsilon^*)\frac{(v-y)}{\lvert v-y\rvert} + y,$$
where $$\varepsilon^* = \begin{cases}\varepsilon &: \lvert v-y\rvert<\varepsilon\\
2\varepsilon - \lvert v-y\rvert &: \varepsilon\leq \lvert v-y\rvert<2\varepsilon \\ 0&:2\varepsilon\leq \lvert v-y\rvert.\end{cases}$$

Consider that $f_y$ is continuous on $\mathbb{R}^d\setminus\{y\}$. And since $y\notin H(\mathbb{D})$, it follows that $f_y\circ H$ is continuous. Also $f_y$ is the identity on $\mathbb{R}^d\setminus B_{2\varepsilon}(x_n)$; so the fact that $B_{2\varepsilon}(x_n)$ is disjoint from $\gamma(S^1)$ implies that $(f_y\circ H)\rvert_{\partial \mathbb{D}} = f_y\circ \gamma=\gamma$, and so we have that $f_y\circ H \in \mathcal{H}$.

Also note that $\lvert f_{y}(v)-v\rvert \leq \varepsilon$ for every $v\in \mathbb{R}^d\setminus\{y\}$, and so $$\lVert(f_y\circ H) - H\rVert_\infty \leq \varepsilon.$$ Thus the triangle inequality yields
$$\lVert(f_y\circ H) - h\rVert_{\infty} \leq 3\varepsilon.$$

Finally, it is clear that the open ball $B_\varepsilon(y)$ is disjoint from the image of $f_y$, and so $B_\varepsilon(y)$ must be disjoint from $(f_y\circ H)(\mathbb{D})$. But $x_n\in B_\varepsilon(y)$, so we have that $x_n\notin (f_y\circ H)(\mathbb{D})$, which shows that $(f_y\circ H)\in S_n$.

Thus it has been shown that each $S_n$ is open and dense in $\mathcal{H}.$ And because $\mathcal{H}$ is a Baire-space, the BCT implies that $\bigcap_{n\in\mathbb{N}} S_n\neq \emptyset$. Any $h\in \bigcap_{n\in\mathbb{N}} S_n$ is a homotopy in $\mathbb{R}^n\setminus \{x_1,x_2,\dotsc\}$ from $\gamma$ to a point.

NOTES: as to @JoelDavidHamkins's comment, this argument clearly generalizes for $\{x_1,x_2,\dotsc\}$ replaced by any set of cardinality less than $\operatorname{cov}(\mathcal{M})$.

This argument does not explicitly use the fact that $\mathbb{R}^d$ minus a finite set is simply connected, but it does use the fact that, the linear image of a $(d-1)$-dimensional polygon has no interior in $\mathbb{R}^d.$

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