To follow up on the answer of @VARulle let me give you some conditions for which the indifference curve is path connected.
The argument can also be found in the book Mathematical Methods and Models for Economists by Angel de la Fuente.
Preferences are monotone if $x > y$ implies $x \succ y$ and that preferences are continuous if $x_n \succeq y_n$, $x_n \to x$ and $y_n \to y$ imply $x \succeq y$.
Let us limit ourselves to the space $\mathbb{R}^n_{++}$ of strictly positive bundles. For some $a \in \mathbb{R}^n_{++}$ consider the indifference curve $I(a) = \{x \in \mathbb{R}^n_{++}| x \sim a\}$.
Theorem If preferences are monotone and continuous then for $a \in \mathbb{R}^n_{++}$, the set $I(a)$ is path-connected.
As being path-connected is a topological property, it suffices to show that there is a homeomorphism between $I(a)$ and a path-connected set. For this, we will take the set $\Delta$:
$$
\Delta = \{z \in \mathbb{R}^n_{++}: \sum_i z_i = 1\}
$$
This set is path-connected as any two elements in $\Delta$ can be connected by the line segment that has these points as endpoints.
Consider the radial projection function $f: I(a) \to \Delta$ defined as:
$$
f(x) = \dfrac{x}{\sum_i x_i}
$$
As we restrict ourselves to $\mathbb{R}^n_{++}$, the function $f$ is well defined. It is easy to see that $f$ is continuous. To show that it is a homeomorphism, we have to show that it is a bijection and that $f^{-1}$ is also continuous.
First, to show that $f$ is onto. let $z \in \Delta$ and consider the ray $\alpha z$ with $\alpha > 0$. Then for $\alpha$ large enough, we have $\alpha z > a$, so by monotonicity $\alpha z \succ a$. Also for $\alpha > 0$ small enough, we have $\alpha z < a$ so, by monotonicity, $a \succ \alpha z$.
Let $\alpha^\ast = \inf \{\alpha| \alpha z \succ a\}$ which is well defined by the argument above. Then by continuity $\alpha^\ast z \succeq a$. Also for all $\alpha < \alpha^\ast$ we have $a \succeq \alpha z$ so again by continuity $a \succeq \alpha^\ast z$. This shows that $\alpha^\ast z \sim a$ or equivalently $\alpha^\ast z \in I(a)$. Defining $x = \alpha^\ast z$, gives that $x \sim a$ and $h(x) = z$, so $h$ is onto.
To show that $f$ is a bijection, let $f(x) = f(y) = z$. We need to show that $x = y$. We have:
$$
\dfrac{x}{\sum_i x_i} = \dfrac{y}{\sum_i y_i} = z.
$$
As such,
$$
x \sum_i y_i = y \sum_i x_i,
$$
which shows that $x$ and $y$ are proportional to each other, i.e.,
$$
\alpha x = \beta y.
$$
if $\alpha = \beta$ then $x$ and $y$ are identical, so either $\alpha > \beta$ or vice versa. But then by monotonicity $y \succ \frac{\beta}{\alpha} y = x$ or vice versa, which gives a contradiction with $x, y \in I(a)$.
Finally, to see that $f^{-1}$ is continuous, let $z \in \Delta$ and take any sequence $z_n \to z$. It suffices to show the existence of a subsequence $z_{k_n}$ such that $h^{-1}(z_{k_n}) \to h^{-1}(z)$.
Define $x_n = h^{-1}(z_n)$ and $x = h^{-1}(z)$. Then:
$$
\alpha_n x_n = z_n.
$$
with $\alpha_n = \frac{1}{\sum_i x_{i,n}}$ also we have that:
$$
\alpha x = z,
$$
with $\alpha = \frac{1}{\sum_i x_i}$. It suffices to show that there is a subsequence for which $x_{k_n} \to x$.
If $\{\alpha_n\}$ is bounded, there is a convergent subsequence $\alpha_{k_n} \to \alpha^\ast$. If $\alpha^\ast = 0$ then as $z_n \to z$ there is an $n_k$ large enough such that $x_{n_k} = \frac{z_{n_k}}{\alpha_{n_k}} > \frac{z}{\alpha} = x$ which would contradict $x_{n_k} \sim x$. As such, $\alpha^\ast > 0$ and:
$$
x_{k_n} = \frac{z_{k_n}}{\alpha_{k_n}} \to \frac{z}{\alpha^\ast} = \frac{\alpha}{\alpha^\ast} x.
$$
Then as $x_{k_n} \sim x$ for all $n$, by continuity, $x \sim \frac{\alpha}{\alpha^\ast}x$ which can only hold (by monotonicity) if $\alpha^\ast = \alpha$. This shows that $x_n \to x$.
If $\alpha_n > 0$ is unbounded, then there is a subsequence $\alpha_{k_n} \to \infty$. Then along this subsequence (as $z_{k_n} \to z$) we have
$$
x_{k_n} = \frac{z_{k_n}}{\alpha_{k_n}} \to 0,
$$
So for $n$ large enough $x_{k_n} < x$ which means that $x \succ x_{k_n}$ which again contradicts $x_{k_n} \sim x$. This shows that $f^{-1}$ is continuous $\square$.
