Proof:

The second part is obvious. To prove the first part, for any loop $\gamma$ in $X$ based at $x_0$,

$$(g \circ f) \circ \gamma = g \circ (f \circ \gamma)$$

so we get upon passing to equivalence classes,

$$(g \circ f)_{*} [\gamma] = g_{*} [f\circ\gamma] = g_{*}(f_{*}([\gamma]))$$

In particular if $f : X \longrightarrow Y$ is a homeomorphism then $f_{*} : \pi_1(X, x_0) \longrightarrow \pi_1(Y, y_0)$ is an isomorphism of groups.$\square$