Something which is generally true is that a homotopy equivalence $\alpha: X\to Y$ induces isomorphisms on all homotopy groups. That is, $\alpha_*: \pi_n(X) \to \pi_n(Y)$ is an isomorphism for all $n$. In the case $n=0$, $\pi_0$ is just a set, not a group, and $\alpha_*$ is a bijection. Thus $X$ is path connected iff $\#\pi_0(X) = 1 \iff \#\pi_0(Y) = 1$ iff $Y$ is path connected.
I haven't told you why this is true though, and really, that is what I should do. Given $y_0, y_1 \in Y$, consider $\beta(y_0), \beta(y_1) \in X$, and let $\gamma:I\to X$ be a path from $\beta(y_0)$ to $\beta(y_1)$. Then $\alpha \circ \gamma:I \to Y$ is a path from $\alpha\circ\beta(y_0)$ to $\alpha\circ\beta(y_1)$. Additionally, the homotopy $H: Y\times I \to Y$ from $\alpha\circ \beta$ to $id_Y$ defines a path from $\alpha\circ\beta(y_0)$ to $y_0$, and likewise for $y_1$. Concretely, the path is $t\mapsto H(y_0, t)$. Thus, we may compose these paths to get one from $y_0$ to $y_1$.