In my general topology textbook there is the following exercise:
Let $m, c \in \mathbb R$ and $X$ the subspace of $\Bbb R^2$ given by $X=\{(x,y) \in \mathbb R^2:y=mx + b\}$. Prove that $X$ is homeomorphic to $\Bbb R$.
I came up with a proof for this, but I think I over-complicated it, nonetheless I still want to know if it is correct or not.
My proof:
Let's redefine $X$ as: $X = \{(t,mt+c):t \in \Bbb R\}$. Now we can define the following funcion:
$$f:\Bbb R \to X$$ $$f(x)=(x,mx + c)$$
This function is an bijection. Now we shall prove that $f$ is continuous. Let $\mathcal B_{X}$ denote the basis for the topological space $(X,\tau_X)$. Let $\mathcal B$ be the basis for $(\mathbb R,\tau)$ and $\mathcal B'$ the basis for $ (\mathbb R^2,\tau')$.
Let $A \in \tau_X$, then we have that, for some index set $I$, $A = \bigcup\limits_{i \in I} B_i$, with $B_i \in \mathcal B_X$.
So we have that: $$f^{-1}(A) = \bigcup_{i \in I} f^{-1}(B_i)$$
Let's define $S_{a \to b}^{c \to d} := \{(x,y) \in \mathbb R^2 : a < x < b \text{ and } c < y < d\}$. Then we have that $\mathcal B' = \{S_{a \to b}^{c \to d}: a,b,c,d \in \mathbb R\}$, so because $(X,\tau_X)$ is a subspace of $\mathbb R^2$ we have that for each $i$:
$$B_i = S_{a_i \to b_i}^{c_i \to d_i} \cap X $$
For some $a_i, b_i, c_i, d_i$.
So we have:
$$\bigcup_{i \in I} f^{-1}(B_i) = \bigcup_{i \in I} f^{-1}( S_{a_i \to b_i}^{c_i \to d_i} \cap X )$$
We have that $f^{-1}( S_{a_i \to b_i}^{c_i \to d_i}\cap X ) = (\alpha_i, \beta_i) \subset \mathbb R$, for some $\alpha_i, \beta_i \in \mathbb R$:
So we have that: $$\bigcup_{i \in I} f^{-1}( S_{a_i \to b_i}^{c_i \to d_i} \cap X ) = \bigcup_{i \in I}\ (\alpha_i,\beta_i)$$
Because each $(\alpha_i,\beta_i) \in \tau$, then $\bigcup\limits_{i \in I}\ (\alpha_i,\beta_i) \in \tau$, so we have that $f$ is continuous.
Now let $A \in \tau$, then we have, for some index set $J$, that $A = \bigcup\limits_{j \in J} \ (\alpha_j , \beta_j)$ for $(\alpha_j , \beta_j) \in \tau$.
$$f(A) = \bigcup_{j \in J} f((\alpha_j , \beta_j))$$
Because $f^{-1}( S_{a_j \to b_j}^{c_j \to d_j} \cap X ) = (\alpha_j, \beta_j)$, then for all $(\alpha_j, \beta_j):$ $$f((\alpha_j, \beta_j)) = S_{a_j \to b_j}^{c_j \to d_j} \cap X$$
So we have that:
$$\bigcup_{j \in J} f((\alpha_j , \beta_j)) = \bigcup_{j \in J} (S_{a_j \to b_j}^{c_j \to d_j} \cap X)$$
Because $(S_{a_j \to b_j}^{c_j \to d_j} \cap X) \in \mathcal B_X \subset \tau_X$, then we have that $\bigcup_{j \in J} (S_{a_j \to b_j}^{c_j \to d_j} \cap X) = f(A) \in \tau_X$, so we have that $f^{-1}$ is continuous.
So there exists $f: \mathbb R \to X$ such that $f$ is bijective, continous and $f^{-1}$ is continous, this $\mathbb R \cong X$
So my question is, is this proof correct? What can I do to improve it? Is there a more straightforward way Of proving this?
Just so this question has an answer, let's sum up the comments. Your proof is correct, and in a broad sense it's as simple as it gets. However, the main point that makes your proof longer is that although you know that $$f^{-1}\left(\bigcup_{i\in I} B_i\right) = \bigcup_{i\in I}f^{-1}(B_i)$$ and you use this crucial fact in your proof, you continue to carry around unions of basis elements. You could simply prove, once, that because of this identity, it is enough to show that $f^{-1}(B)$ is open for each basis element $B$ (conversely, $f(B)$ is open for a basis element in the domain space as well).