Is there an injection from the set of all real sequences to R? Is there an injection from the set of all real sequences
to R?
 A: Yes. In fact there is an easy bijection, not that explicit, but it works.
Let $f: \mathbb{N}^\mathbb{N} \to \mathbb{R}$ be a bijection (there are many, for instance continued fractions). The set of real sequences is $\mathbb{R}^\mathbb{N}$, that corresponds, from the above bijection, to the set of sequences of "sequences of naturals" $(\mathbb{N}^\mathbb{N})^\mathbb{N}$.
But this last set is easily in bijective correspondence to $\mathbb{N}^{\mathbb{N}^2}$, the set of double-sequences of naturals. Finally, $\mathbb{N}^2$ has a bijection to $\mathbb{N}$, which you can chain again to obtain
$$
\mathbb{R}^\mathbb{N} \simeq
(\mathbb{N}^\mathbb{N})^\mathbb{N} \simeq
\mathbb{N}^{\mathbb{N}^2} \simeq
\mathbb{N}^\mathbb{N} \simeq \mathbb{R}
$$
A: Yes. You can show there is a bijection with the following theorem.

If $a$ is an infinite cardinal number, and if $b$ is a cardinal number
  with $2 ≤ b ≤ 2^a$, then $b^a = 2^a$.

Also recall that $|\mathbb{R}| = \beth_1$ and $|\mathbb{N}|=\beth_0$. See also, beth numbers. Thus we may argue as follows. $$|\mathbb{R}^\mathbb{N}| = {\beth_1}^{\beth_0} = 2^{\beth_0} = \beth_1$$
We deduce that $|\mathbb{R}^\mathbb{N}|=|\mathbb{R}|$, or in other words that these sets can be put into bijective correspondence.
