On sort-of-linear functions: does $f(x+y) = f(x) + f(y)$ imply $f(\alpha x) = \alpha f(x)$?

It is not true that $|ax|=a|x|$; the correct identity is $|ax|=|a||x|$.

Whether or not adding the hypothesis of continuity is necessary for additive functions to be linear depends on the axiom of choice. Using a Hamel basis $B$ for $\mathbb{R}^n$ over $\mathbb{Q}$ together with one of its countable subsets $A=\{x_1,x_2,\ldots\}$, you can construct a discontinuous $\mathbb{Q}$ linear map from $\mathbb{R}^n$ to $\mathbb{R}$ by taking the unique $\mathbb{Q}$ linear extension of the function $f:B\to\mathbb{R}$ such that $f(x_k)=k|x_k|$ and $f(B\setminus A)=\{0\}$. Since $\mathbb{R}$ linear maps between finite dimensional real vector spaces are continuous, such a map cannot be linear. However, it is consistent with ZF that all additive functions from $\mathbb{R}^n$ to $\mathbb{R}$ are continuous (I am however not knowledgeable in the set theoretic background needed to show this).

Yes, continuity is necessary. In one variable the counterexamples are known as pathological solutions to Cauchy's functional equation. They require the axiom of choice to construct: given the axiom of choice, we can pick a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and any function $f : \mathbb{R} \to \mathbb{R}$ which is only required to be $\mathbb{Q}$-linear can be specified by arbitrarily specifying its behavior on this basis.

Pathological solutions are very strange (for example their graph is dense in the plane) and can be ruled out by almost any "niceness" hypothesis, as described in the wiki article.