Given a measurable space $X$ equipped with a measure $\mu$, a function $f : X \to \mathbb{C}$ which is defined almost everywhere (that is, up to a set of measure $0$) is said to be an element of $L^1$ if
$$\int_X |f| d\mu < \infty$$
More properly, we have to identify functions which are equal almost everywhere, so the elements of the Lebesgue space $L^1$ are really equivalence classes of functions under the relation of being almost everywhere equal - but this is a technical note.
Practically speaking, a real or complex valued measurable function on the real line with respect to Lebesgue measure is an element of $L^1$ if $$\int_{-\infty}^{\infty} |f(x)| dx < \infty$$
So a function like $\chi_{[0,1]}$ which is $1$ on $[0,1]$ and $0$ outside is in $L^1$, as is $e^{-x^2}$.
If $f$ is continuous enough, this coincides with the usual Riemann integral. Now it's fairly easy to prove that $$\int_{\mathbb{R}} |\sin x| dx$$ can't be finite, so $\sin x \notin L^1(\mathbb{R})$. In some sense, $L^1$ functions have to decay to $0$ at $\pm \infty$: In fact, one way to think of $L^1$ is that it's the completion of
$$C_C = \{\text{continuous functions supported on a compact set}\}$$
under the metric induced by integration (again, with slight technical caveats).
So in short, ignoring the technical definitions, $L^1$ functions are exactly those functions which have small enough spikes and decay fast enough that $\int |f| < \infty$.
Yes. Your argument is valid. In a nutshell, your argument is this:
Take any $f \in X^*$. Since $X$ is finite dimensional, $f$ is bounded. So, $f \in X'$. So, $X^* \subseteq X'$, as desired.
When $X$ is infinite dimensional, we can always construct an unbounded linear function $f$ on $X$. Finding an explicit construction of such an $X$ is tricky. However, it suffices to find an example that works on a vector space of countably infinite dimension (it is non-trivial to see that this is indeed enough; you'll need to inject the axiom of choice appropriately).
To that end: let $\{e_k\}_{k \in \Bbb N}$ be a (Hamel-)basis for $X$ consisting of unit-vectors. We can the define a linear functional by
$$
f(e_n) = n
$$
Hopefully you can see that this must be unbounded.
Best Answer
The dual space consists of all continuous linear functionals on the space. What continuity buys you is that, if you can determine the functionals on a dense subspace $M$, you can bootstrap to the full space by continuity. Continuity of a linear function $F$ on the normed space assures that $F$ is determined by its values on the dense subspace $M$.
For example, because $\ell^1$ has a basis of sequences $$ e_0 = \{ 1, 0, 0, 0, \cdots \} \\ e_1 = \{ 0, 1, 0, 0, \cdots \} \\ e_2 = \{ 0, 0, 1, 0, \cdots \} \\ \vdots $$ then there is a natural way to bootstrap. Specifically, if $x=\{ \alpha_k \}\in\ell^1$, then $$ \left\| x - \sum_{n=0}^{N}\alpha_n e_n \right\|_{\ell^1} \le \sum_{k=N+1}^{\infty}|\alpha_n|\rightarrow 0 \mbox{ as } N\rightarrow\infty. $$ Therefore, by continuity, if $F$ is a continuous linear functional on $\ell^1$, then $$ F(x) = \lim_{N\rightarrow\infty}F\left(\sum_{k=0}^{N}\alpha_k e_k \right) = \lim_{N\rightarrow\infty} \sum_{k=0}^{N}\alpha_k F(e_k) = \sum_{k=0}^{\infty}\alpha_k F(e_k). $$ So, already you have a representation of any such linear functional. The sum on the right is guaranteed to converge. In fact it must converge absolutely because you can choose unimodular constants $u_k$ such that $u_k\alpha_k F(e_k)=|\alpha_k||F(e_k)|$ and apply the above to $\{u_k\alpha_k\}$ instead. Continuity of $F$ is equivalent to boundedness, meaning the existence of a smallest non-negative constant $\|F\|$ such that $|F(x)| \le \|F\|\|x\|_{\ell^1}$ for all $x\in\ell^1$. From this, you get $\{ F(e_k) \}\in\ell^{\infty}$ and $\|\{ F(e_k)\}\|_{\ell^{\infty}} \le \|F\|$. Conversely, $$|F(\{\alpha_k\})| \le \|\{F(e_k)\}\|_{\infty}\|\{\alpha_k\}\|_{\ell^1}$$ gives $\|F\|\le\|\{F(e_k)\}\|_{\infty}$ because $\|F\|$ is the smallest such constant. Hence, $$\|F\|=\|\{F(e_k)\}\|_{\ell^{\infty}}.$$ So there is an isometric linear correspondence between $(\ell^{1})^*$ and $\ell^{\infty}$.