Existence of an injective continuous function $\Bbb R^2\to\Bbb R$?
Assume $f : \mathbb{R}^2 \to \mathbb{R}$ is continuous and injective. Then for each fixed $y$, the function $x \mapsto f(x,y)$ is monotonic. Its image is some interval, and in particular contains a rational number. None of these points can be re-used for some other $y$. So $y$ can't be drawn from an uncountable set, since the rationals are countable. But $\mathbb{R}$ is uncountable...
If $f$ takes values in $\mathbb R$, then no, $f$ cannot be injective.
To see this, consider the function $$ g(t) = f(\cos t, \sin t) $$ which is continuous because it is a composition of continuous functions. If $f(1,0)=f(-1,0)$ then we know $f$ is not injective. Otherwise, since $g(0)=g(2\pi)$, by the intermediate value theorem the value $\frac{g(0)+g(\pi)}{2}$ must be attained for some $t_0\in(0,\pi)$ as well as for some $t_1\in(\pi,2\pi)$. But $(\cos t_0,\sin t_0)$ is a different point than $(\cos t_1, \sin t_1)$. so in this case $f$ is not injective either.
On the other hand, there are continuous injections $\mathbb Q^2\to\mathbb Q$. Namely, $(x,y)\mapsto x+\sqrt2 y$ injects $\mathbb Q^2$ continuously into $\mathbb Q+\sqrt2 \mathbb Q$, and this set is a countable dense linear order without first and last element and therefore order-isomorphic -- and thus homeomorphic -- to $\mathbb Q$.
No, because the continuous image of a connected set is connected. If $f: \mathbb{R}^{2} \to \mathbb{R}$ is a continuous bijection, then we can consider $f( \mathbb{R}^{2} \setminus \{ 0, 0 \} ) = \mathbb{R} \setminus \{ f(0, 0) \}$. Let $c = f(0, 0)$. Since $\mathbb{R}^{2} \setminus \{ (0, 0) \}$ is connected, so must be $\mathbb{R} \setminus \{ f(0, 0) \} = ( - \infty, c) \cup (c, + \infty)$, which it obviously isn't. Thus either $f$ is not continuous, or it is not a bijection.
EDIT: Commenter pointed out that the OP only supposed $f$ was injective. A similar method holds. Again, $f (\mathbb{R}^{2})$ is a connected subset of $\mathbb{R}$, and so is an interval, since all connected subsets of $\mathbb{R}$ are intervals (though its endpoints may be infinite). So suppose that the interval $(a, b) \subseteq f(\mathbb{R}^{2})$. Let $a < c < b$, and let $\mathbf{x} = (x_{1}, x_{2})$ be such that $f(\mathbf{x}) = c$. Then again, $\mathbb{R}^{2} \setminus \{ \mathbf{x} \}$ is connected, so $f( \mathbb{R}^{2} \setminus \{ \mathbf{x} \} )$ is connected. But we can consider that $$f( \mathbb{R}^{2} \setminus \{ \mathbf{x} \} ) = ( (- \infty, c ) \cap f(\mathbb{R}^{2} \setminus \{ \mathbf{x} \})) \cup ( (c, + \infty) \cap f(\mathbb{R}^{2} \setminus \{ \mathbf{x} \})) ,$$ and so is not connected. Thus either $f$ is not injective, or its not continuous.