Show if $f:\Delta^n\rightarrow X$ is continuous then $f$ is constant

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Let $\prod_{k=0}^{\infty} S^0$. Call $X_{dis}$ $X$ equipped with the discrete topology and $X_{prod}$ $X$ equipped with the product topology. Let $j: X_{dis}\rightarrow X_{prod}$ be the identity.

Show that a continuous function from $\Delta^n$ to either $X_{dis}$ or $X_{prod}$ is necessarily constant. Then show $j_*:H_*(X_{dis})\rightarrow H_*(X_{prod})$ is an isomorphism.

The only thought I had was to use the open set definition of continuity to force the function to be constant but couldn't figure out how to actually make that work.

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Both of the spaces $X_{dis}$ and $X_{prod}$ are totally path disconnected, i.e. every path component is a point.

This is obvious for $X_{dis}$ since every point is an open set.

For $X_{prod}$, this can be proved most directly by considering the $n^{\text{th}}$ composition $X_{prod} \to S^0$ indexed by any integer $n \in \{0,1,2,3,\ldots\}$, and using the fact that $S^0$ is discrete and therefore totally path disconnected. The image of any continuous path in $X_{prod}$ when projected to $S^0$ is a constant (since $S^0$ is totally path disconnected), hence each coordinate function of each path in $X_{prod}$ is constant, hence the whole path is constant.

Once you have that, the next thing you use is the easy fact that any continuous function from any path connected space to any totally path disconnected space is constant.