Consider the subset $X = \{(x, 0) \mid -1 \lt x \lt 1\} \cup \{(0, y) \mid -1 \lt y \lt 1\} \subset \Bbb R^2$. It's just two open segments intersecting at $(0, 0)$.
How do you show that $X$, equipped with the subspace topology of $(\Bbb R^2, \lvert \lvert \cdot \rvert \rvert_2)$, is not locally homeomorphic to an euclidean space because of the point $(0, 0)$ ?
First I'll prove that $X$ is not homeomorphic to $\mathbb{R}^n$
Proof: Suppose $X$ is homeomorphic to $Y =\mathbb{R}^n$ for some $n > 0$. Let $f$ denote such a homeomorphism. $X$ is connected and thus $f[X] = Y$ is connected by continuity of $f$.
Now consider $f' : X \backslash \{0\} \to Y \backslash \{f(0)\}$, where $f'$ is the restriction of $f$ to the domain $X \backslash \{0\}$. Clearly $f'$ is continuous, bijective and has continuous inverse, so it is a homeomorphism between $X \backslash \{0\}$ and $Y \backslash \{f(0)\}$.
Observe that $X \backslash \{0\}$ has 4 connected components (draw a picture to see why this is true), whereas $Y \backslash \{f(0)\}$ has 2 connected components when $n = 1$, and 1 connected component (that being $Y \backslash \{f(0)\}$) when $n \geq 2$.
Since homeomorphisms preserve connected components, we arrive at a contradiction, so $X$ cannot be homeomorphic to $\mathbb{R}^n$ for any $n > 0$. $\square$.
Now I'll prove that $X$ is not locally homeomorphic to any open set of $\mathbb{R}^n$ for $n > 0$. This proof is almost the same as the above proof.
Proof: Suppose $X$ is locally homeomorphic to $\mathbb{R}^n$ for $n > 0$. Let $f : X \to \mathbb{R}^n$ be such a local homeomorphism. Then for each $x \in X$, there exists an open set $U$ containing $x$, such that $f[U]$ is open in $\mathbb{R}^n$ and the restriction $f|_U : U \to f[U]$ is a homeomorphism.
Choose $x =0 \in X$. Then by the definition of a local homeomorphism, there exists an open set $V$ containing $0$ . $V$ is connected, and thus $f[V] \subseteq \mathbb{R}^n$ is also connected by continuity of $f$. We also have $f[V]$ to be open in $\mathbb{R}^n$, and $f|_V : V \to f[V]$ to be a homeomorphism by the definition of a local homeomorphism.
Consider $f' : V \backslash\{0\} \to f[V] \backslash \{f(0)\}$ where $f'$ is the restriction of $f|_V$ to the domain $V \backslash \{0\}$. Clearly $f'$ is continuous, bijective and has continuous inverse (due to the fact that $f|_V$ is a homeomorphism), so it is a homeomorphism between $V \backslash\{0\}$ \to $f[V] \backslash \{f(0)\}$.
Observe that $V \backslash \{0\}$ has 4 connected components (draw a picture to see why this is true), whereas $f[V] \backslash \{f(0)\}$ has 2 connected components when $n = 1$, and 1 connected component (that being $f[V] \backslash \{f(0)\}$) when $n \geq 2$.
Since homeomorphisms preserve connected components, we arrive at a contradiction, so $X$ cannot be locally homeomorphic to $\mathbb{R}^n$ for any $n > 0$. $\square$.
