I'll answer by jotting down a few basic results in quantum information. For convenience let's adopt the notation
$$
| \uparrow \rangle \equiv |0\rangle,\qquad | \downarrow \rangle \equiv |1\rangle .
$$
For example, the singlet state is $(|01\rangle - |10\rangle)/\sqrt{2}$.
Let's also define the single-qubit operators
\begin{eqnarray}
X &\equiv& | 0 \rangle\langle 1 | + | 1 \rangle\langle 0 |,\\
Z &\equiv& | 0 \rangle\langle 0 | - | 1 \rangle\langle 1 |,\\
Y &\equiv& Z X.
\end{eqnarray}
$X$ and $Z$ are Pauli operators, and $Y$ is $i$ times a Pauli operator.
The notation $ZX$ means a product of operators acting on a single qubit.
But we also want to write tensor products of operators acting on different qubits. For that purpose we add a subscript, so for example $Z_1 X_2$ means a $Z$ acting on the first qubit and an $X$ acting on the second qubit.
Now we can notice that the singlet state is an eigenstate of $Z_1 Z_2$ with eigenvalue $-1$:
$$
Z_1 Z_2 \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)
= \frac{1}{\sqrt{2}}(-|01\rangle + |10\rangle)
= -\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)
$$
and it is also an eigenstate of $X_1 X_2$ with eigenvalue $-1$ (exercise for the reader).
How about the state $(|00\rangle + |11\rangle)$? (I am not bothering with normalization now). It is an eigenstate of $Z_1 Z_2$ and of $X_1 X_2$ with eigenvalue $1$ in each case.
You can now explore the other Bell states.
The thing to notice is, of course, that these entangled states are not eigenstates of any single-qubit operator. So to measure a state and be sure that after the measurement the state is entangled, you have to measure two different two-qubit observables, in such a way that you do not also measure any single-qubit observables whose measurement would destroy the entanglement. The experimental method to measure an observable such as $Z_1 Z_2$ without measuring either $Z_1$ or $Z_2$ on its own
is whatever a good experimentalist would like to propose, but the following is one good way.
To measure $Z_1 Z_2$ we may combine two-qubit operators with single-qubit measurements. To measure $Z_1 Z_2$ (without measuring either of $Z_1$ or $Z_2$ on its own) you can, for example, perform controlled-not from each qubit to a third qubit (prepared initially in $|0\rangle$) and then measure that third qubit. To measure $X_1 X_2$ you can use that $X = H Z H$ where $H$ is the Hadamard transform. It follows that you can adopt the same method as for $Z_1 Z_2$ but with $H$ operators suitably inserted.
All this is a few notes on basic ideas in quantum information which could also be found in any introductory course notes or textbook.
Postscript
Finally, let me take issue with one statement in the question, where it says "In order to be 100% sure to have an entangled spin state, one would have to measure it". This is first a bit muddled, because if you have a single system in a state that is unknown to you, then performing a measurement can never give 100% certainty about what state it was in before the measurement was enacted, no matter what kind of state or observable is involved. This is because if the state was unknown then you cannot be 100% sure that your measurement did not disturb it. So the only way to get the kind of certainly one is asking for here is either to prepare the state yourself, or to have many systems which some oracle has guaranteed are all in the same state. The former is the way to do it. So to be 100% sure to have an entangled spin state, one would first prepare two spins in a known unentangled state by projective measurement, and then drive them through a known, controlled, unitary evolution to the entangled state you wanted to prepare.
