A question about continuous curves in $\mathbb{R}^2$ Let $f:[a,b]\longrightarrow\mathbb{R}^2$ be a continuous function such that 
$$f(a)=(0,0),\ f(b)=(0,1).$$
Is it true that there must exist $t_1,t_2\in [a,b]$ such that
$\displaystyle f(t_1)-f(t_2)=(0,\frac{1}{2})?$
If not, please give a counterexample.
 A: More of a comment, the particular case $f(t) = (f_1(t), t)$, the graph of a function, is relevant, and somehow well known. In this case, if $f_1(0)=f_1(1)$, there exists $t$, $t+\frac{1}{2}\in [0,1]$ so that 
$f_1(t) = f_1(t+\frac{1}{2})$. The proof should be clear. It should work with $\frac{1}{n}$ -but not with other distances! ( there are counter examples).So for those  proofs that work for any $\delta >0$, they shouldn't. 
${Added:}$ Here is a link to an article with more  relevant references on "horizontal chords in graphs". 
$\bf{Added:}$ Here is an example of a function $f(t) = (f_1(t), t)$ from $[0,1]$ to $\mathbb{R}^2$, $f(0) = (0,0)$, $f(1)= (0,1)$, so that there does not exist on this curve a vertical chord or length $\frac{\pi}{4}$. Take $f_1(t) = \sin(8 t) - t \sin 8$. 
A: This answer is wrong, as pointed out by Wen. I won't delete it: maybe someone will find it useful to create a working counterexample
I think this should work as a counterexample.
Take $a=0$, $b=7$. We will define $f$ piecewisely.
For $t\in[0,1]$,  $\ \ f(t)=(\frac{t}{2},0)  $;
for $t\in[1,2]$, $\ \ f(t)=(\frac{1}{2},\frac{t-1}{4})$;
for $t\in[2,3]$, $\ \ f(t)=(\frac{1}{2}-(t-2),\frac{1}{4})$;
for $t\in[3,4]$  $\ \ f(t)=(-\frac{1}{2},\frac{1}{4}+\frac{3}{8}(t-3))$;
for $t\in[4,5]$, $\ \ f(t)=(-\frac{1}{2}+\frac{3}{2}(t-4),\frac{5}{8})$;
for $t\in[5,6]$, $\ \ f(t)=(1,\frac{5}{8}+\frac{3}{8}(t-5))$;
for $t\in[6,7]$, $\ \ f(t)=(1-(t-6),1)$.
Since it is possible that I made some mistake in the parametrization, the idea is the following:
Start from $(0,0)$; go along the x-axis to the point $(\frac{1}{2},0)$; go up to the point $(\frac{1}{2},\frac{1}{4})$; now "turn back" to the point $(-\frac{1}{2},\frac{1}{4})$; now go up to the point $(-\frac{1}{2},\frac{5}{8})$. At this point you are above the points with y-coordinate $\frac{1}{2}$.  To add $\frac{3}{8}$ vertically, without having two points $t_1,t_2$ s.t. $f(t_1)-f(t_2)=(0,\frac{1}{2})$, you need to go back "beyond" $\frac{1}{2}$ in the x-direction, then you can reach the points with y-coordinate 1 "safely".  
 
A: I'll use the well-known concept of Brouwer topological degree. Here is an brief intuitive idea (roughly speaking) of the construction of such tool..
Given a (at least) $C^2$ function $g\colon\Omega\subseteq\mathbb R^n\to\mathbb R^n$ (or, more generally, between two oriented manifolds of same dimension) and $y\in\mathbb R^n$ a regular value for $g$ such that $g^{-1}(y)$ is compact (therefore finite, by the inverse function theorem), one defines the Brouwer degree of $g$ at $y$ as the integer number
$$\deg(g,y) = \sum_{x\in g^{-1}(y)} {\rm sign}\det g'(x),$$
where $g'(x)$ is the jacobian matrix of $g$ at $x$ and ${\rm sign}\ t$ is $1$ for $t>0$ or $-1$ for $t<0$.
One can prove that this number is homotopy invariant, that is, if $f$ and $g$ are homotopic and $y$ is a regular value for both $f$ and $g$, then
$$\deg(f,y) = \deg(g,y).$$
Using Sard's theorem to define the degree of maps at critical points, we have that, given a $C^2$ function $h$, $\deg(h,\cdot)\colon\mathbb R^n\to\mathbb Z$ is a continuous locally constant function.
Now, given a continuous function $f\colon\Omega\to\mathbb R^n$, we can choose a smooth function $g$ "close enough from $f$" (using Weierstrass approximation theorem) and define the degree of $f$ at some value $y$ by
$$\deg(f,y) = \deg(g,y),$$
where the right side is already defined.
Briefly, the Brouwer degree can be thought as a map taking a continuous function $f\colon\Omega\to\mathbb R^n$ and some value $y\in\mathbb R^n$ and giving a integer number.
In addition, notice that if $\deg(f,y) \neq 0$ then the equation $f(x) = y$ has at least one solution.

Now we're ready to think about the problem.
Let ${\rm I} = [0,1]$ and $q=(0,1/2)$. Consider $\alpha\colon{\rm I}\to\mathbb R^2$ be a continuous function such that $\alpha(0) = (0,0)$ and $\alpha(1) = (0,1)$ and .
Let $\gamma\colon{\rm I}\to\mathbb R^2$ be the smooth function $$\gamma(t) = (\sin(\pi t), t).$$
Here is the plot of $\gamma$:

Define $F, G\colon {\rm I}^2\to\mathbb R^2$ (note the same dimension) by
$$F(x,y) = \alpha(x) - \alpha(y)$$ and $$G(x,y) = \gamma(x) - \gamma(y).$$
We have that $F$ and $G$ are homotopic since it assumes values on $\mathbb R^2$ (we can take the linear homotopy).
Therefore, $\deg(F,q) = \deg(G,q)$.
Let's calculate $\deg(G,q)$. First, we have to verify that $q$ is in fact a regular value for $G$ to then apply the degree formula. We are going to take a look at the set $G^{-1}(q)$.. with some straight forward calculations, we can see that
$$G(x,y) = q  \Leftrightarrow \left\{
\begin{matrix} x = k + 3/4 \\ y = k + 1/4 \end{matrix}\right. , k\in\mathbb Z$$
and $k=0$ gives the solution $(x,y)=(3/4,1/4)$, that lies in ${\rm I}^2$.
So $G^{-1}(q) = \{(3/4,1/4)\}$.
Note that
$$G'(x,y) = \begin{bmatrix}
\pi\cos(\pi x) & -\pi\cos(\pi y) \\
1 & -1 \end{bmatrix},$$
thus
$$\det G'(x,y) = -\pi(\cos(\pi x) - \cos(\pi y))$$
and
$$\det G'(3/4,1/4) = \pi\sqrt{2} > 0.$$
Therefore,
\begin{equation*}
\begin{split}
\deg(G,q) & = \sum_{(x,y)\in G^{-1}(q)} {\rm sign} \det G'(x,y) \\
& = {\rm sign} \det G'(3/4,1/4) \\
& = {\rm sign} \pi\sqrt{2} \\
& = 1.
\end{split}
\end{equation*}
Since $\deg(F,q) = \deg(G,q) = 1 \neq 0$, we conclude that there exists some $(x_0,y_0)\in{\rm I}^2$ such that $F(x_0,y_0) = q$, i.e.,
$$\alpha(x_0) - \alpha(y_0) = (0,1/2).$$
A: This is more like a comment extending on an observation in orangeskid's answer which is also discussed in his linked paper (Proposition 4). It turns out that any (non-zero) vertical chord length can be avoided except $1/n$ with $n\in\Bbb N$. This means that any proof of the initial statement cannot use a general argument that fits any vertical chord length.
Check orangekid's answer to see how the folowing functions are used. Let $$f(x)=\sin(2\pi/\omega\cdot x)-x\sin(2\pi/\omega).$$
Note that $f(0)=f(1)=0$. Consider $g(x):=f(x-\omega)$. We have
\begin{align}
g(x)&=\sin(2\pi/\omega\cdot x-2\pi)-(x-\omega)\sin(2\pi/\omega)\\
&=\sin(2\pi/\omega\cdot x)-x\sin(2\pi/\omega)+\omega\sin(2\pi/\omega)\\
&=f(x)+\omega\sin(2\pi/\omega).
\end{align}
Clearly $f$ and $g$ never intersect except when $\omega=1/n$ for $n\in\Bbb N$. If they do not intersect then this is equivalent to say that the curve $(f(t),t)$ has no (vertical) chord of length $\omega$.
The paper shows (using the intermediate value theorem on the slope of chords) that OP's statement (generalized to chord lengths $1/n$) is indeed true for curves given via $(f(t),t)$ for some function $f:[0,1]\to\Bbb R$ with $f(0)=f(1)=0$ like the one above.
A: I just came across a curious theorem in Rolfsen's book "Knots and Links", and was reminded of this question.

Theorem B.15 (p. 16). Suppose $X$ is a path-connected subset of $\Bbb R^2$, and $C$ is a chord (that is, a line-segment) with endpoints in $X$. Suppose $0<\alpha<1$. Then among all chords with endpoints in X and parallel to $C$, there is either one of length $\alpha|C|$ (where $|C|$ is the length of the chord $C$) or one of length $(1-\alpha)|C|$.

Let $X$ be the curve in your question. Let $C$ be the chord from $(0,0)$ to $(0,1)$. Set $\alpha=1/2$. Theorem B.15 yields the existence of a chord with end points $x,y\in X$ and $x-y=(0,1/2)$.
