 Solution

True. For $u,v\in f(M)$ choose preimages $x,y\in M$. A path $\gamma :[0,1]\to M$ joining $x$ to $y$ exists because $M$ is path-connected. The composition of continuous functions $f\circ \gamma$ is a path in $f(M)$ from $u$ to $v$.Surjectivity onto a larger target is unnecessary because the conclusion concerns the image itself.