Recall that Dirichlet showed the following:
For every real number $x$ and every $Q>1$, there exists an integer vector $(p,q)\in \mathbb Z^2$ such that $|xq-p|<1/Q$ and $0<q<Q$.
I wonder if the following is true:
For every real number $x$ and every $\epsilon>0$, there exists $Q_{\epsilon}$ such that for all $Q>Q_{\epsilon}$, there is an integer vector $(p,q)\in \mathbb Z^2$ such that $|xq-p|<\epsilon/Q$ and $0<q<Q$.
Of course this is trivially true when $x$ is rational, but I don't know what happens when $x$ is irrational (in particular when it is transcendental).
Note the Lerendre's theorem should be a special case of this by taking $\epsilon=1$ and we can always assume $Q>Q_{\epsilon} \ge 1$.
Update: As per the comments by rtybase below, by Liouville's theorem (on diophantine approximation), my statement is false if $x$ is algebraic. But what happens when $x$ is transcendental?
Your conjecture is false. The best possible constant is $\epsilon = \frac 1 {\sqrt 5}$, which is tight for $\phi=\frac{1+\sqrt 5}2$ and its equivalent numbers under the equivalence relation:
$$y\sim x \iff y=\frac{ax+b}{cx+d}; ad-bc=1$$
You can prove that by showing that the best approximants are the convergents to the continued fraction, then noting that those for $\phi$ are ratios of Fibonacci numbers. Then proving that
$$\lim_{n\to\infty}F_n|\phi F_n-F_{n+1}|\to \frac 1{\sqrt 5}$$
Using the closed form for Fibonacci numbers.
If you exclude all the $x\sim\phi$ you can improve the constant to $\epsilon=2\sqrt 2$, and if you repeat the process you find the Lagrange numbers, which converge to 3.
As noted in the comments, the irrationality measure of all irrational algebraic numbers is 2 (Roth's theorem). As for transcendental numbers, almost all of them have irrationality measure 2 still.