$f:X\to Y$ and $X$ is contractible. Let $X_0$ and $Y_0$ be subsets of $X$ and $Y$ respectively. $f_{|X_0}$has values in$Y\setminus Y_0$ and is not null homotopic in $Y\setminus Y_0$. Show that $f(X)\cap Y_0\neq \emptyset$
I would be thankful for hint how to prove it.
I assume you mean $f(X) \cap Y_0 \neq \emptyset$, since your assumption is that $f\mid_{X_0}$ has values in $Y \backslash Y_0$.
If $X$ is contractible, $f:X \rightarrow Z$ is continuous, and $X_0 \subset X$, then $f \mid_{X_0}$ is null homotopic in $Z$. If $f(X) \cap Y_0 = \emptyset$, using this fact with $Z = Y \backslash Y_0$ implies $f\mid_{X_0}$ is null homotopic in $Y \backslash Y_0$, a contradiction.
To prove that assertion: $X$ being contractible means for a point $p \in X$ the map $\pi:X \rightarrow X$, $\pi(x) = p$ is homotopic to the identity. Let $i:X_0 \hookrightarrow X$ denote the inclusion. Then $f \circ \pi \circ i$ is homotopic to $f \circ \text{Id}_{X} \circ i$. But $f \circ \text{Id}_{X} \circ i = f\mid_{X_0}$, and $f \circ \pi \circ i$ is the constant map $x \mapsto f(p)$, so $f \mid_{X_0}$ is homotopic to a constant map, i.e. it is null homotopic.